Can you use Chevalley‒Warning to prove existence of a solution?

Let $f_i \in \mathbb{F}_q[X_1,\dots,X_n]$ have degree at most $d$ for each $i \in [|1,r|]$, and assume that the affine scheme $$ X = \operatorname{Spec} \mathbb F_q [x_1,\ldots,x_n]/(f_1, \ldots, f_r) $$ is nonempty. Take a prime number $\ell \neq p$. Consider the Frobenius $F$ as an endomorphism of the vector space $$ V = \bigoplus_{i \in \mathbb{Z}} H_c^i(X_{\bar{\mathbb{F}_q}}, \mathbb{Q_{\ell}}) $$ By Theorem 1 of this paper by Katz, the dimension of $V$ is bounded above by an explicit constant $B = B(n,r,d)$. Hence one can find elements $\lambda_1,\dots,\lambda_B \in \mathbb{Q}_{\ell}$ such that one has $$ F^{B+1} = \lambda_1 F + \lambda_2 F^2 + \dots + \lambda_B F^B $$ on $V$. It follows that for any $k > B$, the integer $$ |X(\mathbb{F}_{q^k})| = \sum_{i \in \mathbb{Z}} (-1)^i \mathrm{Tr}\left( F^k | H_c^i(X_{\bar{\mathbb{F}_q}}, \mathbb{Q_{\ell}})\right) $$ is a linear combination of $|X(\mathbb{F}_{q})|,|X(\mathbb{F}_{q^2})|,\cdots,|X(\mathbb{F}_{q^B})|$. Since $|X(\mathbb{F}_{q^k})|$ is nonzero for some $k$, it must be nonzero for some $k \leq B$. Hence this constant $B$ yields a positive answer to your question.

The answer by js21 is a great answer, and it gives an explicit choice of $B$. However, you have not asked for any restrictions on $B$ other than it depends only on $n$, $r$, and $d_1,\dots,d_r$ (I assume you want it to be independent of $q$ and of the coefficients of $f_1,\dots,f_r$). There is, in fact, such a $B$ for every choice of $n$, $r$, and $d_1,\dots,d_r$, even if $r>n$, and this only uses the basic techniques of existence of the Hilbert scheme. Over $\text{Spec}(\mathbb{Z})$, you can form the following parameter space as a product of projective spaces $$\Pi = \mathbb{P}H^0(\mathbb{P}^n,\mathcal{O}(d_1))\times \dots \times \mathbb{P}H^0(\mathbb{P}^n,\mathcal{O}(d_r)).$$ Inside the product $\Pi\times \mathbb{P}^n$, you can form the universal closed subscheme $Z$ that is the simultaneous zero locus of an $r$-tuple of homogeneous polynomials of degrees $d_1,\dots,d_r$.

The scheme $\Pi$ is finite type over $\text{Spec}(\mathbb{Z})$, and the projection morphism $\pi:Z\to \Pi$ is projective. Thus, only a finite number of Hilbert polynomials may occur for geometric fibers of $\pi$. More precisely, for every scheme-theoretic fiber of $\pi$ over a point, and for every nonempty open subset of the fiber, we may consider the scheme-theoretic closure of that open subset as a closed subscheme of the fiber. There are only finitely many Hilbert polynomials that can occur for all of these schemes. This follows quicky from the existence of a flattening stratification of $\pi$.

So now let $\mathcal{E}=(e_1(t),\dots,e_N(t))$ be a finite set of numerical polynomials. We say that $\mathcal{E}$ is "saturated" if for every field $F$, both (a) for every $e(t)$ in $\mathcal{E}$, the first difference $e(t) - e(t-1)$ is also in $\mathcal{E}$, and (b) for every closed subscheme $Y\subset \mathbb{P}^n_F$ whose Hilbert polynomial is in $\mathcal{E}$, for every nonempty open subset of $Y$ with its reduced structure, the scheme-theoretic closure of that open set also has Hilbert polynomial in $\mathcal{E}$. By the existence of the Hilbert scheme with fixed Hilbert polynomial as a finite type scheme, by the existence of the flattening stratification, etc., every finite $\mathcal{E}$ is contained in a finite $\mathcal{E}$ that is saturated with respect to (b). It is straightforward to saturate with respect to first differences without altering $\mathcal{E}_m$. Thus, by induction on $m$, every $\mathcal{E}$ is contained in a finite $\mathcal{E}$ that is saturated.

We can partition $\mathcal{E}$ as $\mathcal{E}_0\sqcup \dots \sqcup \mathcal{E}_m$, where $\mathcal{E}_k \subset \mathcal{E}$ is the subset of numerical polynomials of degree $k$. Fix a hyperplane $H=\mathbb{P}^{n-1}\subset \mathbb{P}^n$, i.e., a "hyperplane at infinity" whose complement is an affine space $\mathbb{A}^n$. The claim is that there exists an integer $d=d(\mathcal{E})$ such that for every field $F$ (e.g., a finite field), for every closed subscheme $Y\subset \mathbb{P}^n_F$ whose Hilbert polynomial is in $\mathcal{E}$ and such that $Y$ is not contained in $\mathbb{P}^{n-1}_F$, then there exists a field extension $F'/F$ of degree $\leq d$ such that $Y(F')$ contains an element that is not in $\mathbb{P}^{n-1}(F')$. Up to replacing $Y$ by one irreducible component $Y_j$ that is not contained in $\mathbb{P}^{n-1}_F$ (this is allowed since $\mathcal{E}$ is saturated), we may assume that $Y$ is integral and $Y\cap \mathbb{P}^{n-1}_F$ is the zero scheme of a regular section of $\mathcal{O}_Y(1)$. In particular, the Hilbert polynomial of this zero scheme is the first difference of the Hilbert polynomial of $Y$, thus it is also in $\mathcal{E}$ since $\mathcal{E}$ is saturated.

The claim is proved by induction on $m$. The base case is when $m$ equals $0$. In this case, every $e_i(t)$ is a constant polynomial, and we let $d$ be the maximum of these integers. Since the Hilbert polynomial of $Y$ is constant, $Y$ is zero-dimensional. Then $Y_j$ has total degree $\leq d$. Thus, the associated reduced scheme of $Y_j$ has total degree $\leq d$. This reduced scheme is a singleton set of a closed point that is not contained in $\mathbb{P}^{n-1}_F$. Thus, the residue field of this closed point gives $F'/F$ of degree $\leq d$. This proves the claim when $m$ equals $0$.

By way of induction, assume that $m>0$ and assume the result has been proved for smaller $m$. By the previous paragraph, there exists an integer $d$ that works for $\mathcal{E}_0$. Thus, without loss of generality, assume that the Hilbert polynomial of $Y$ has degree $>0$, i.e., $Y$ has dimension $>0$. Since the Hilbert polynomial of $Y\cap \mathbb{P}^{n-1}_F$ is in $\mathcal{E}$, and since $\mathcal{E}$ is saturated, every irreducible component $D_j$ of $Y\cap \mathbb{P}^{n-1}_F$ has Hilbert polynomial in $\mathcal{E}$. The number $\ell$ of such components is bounded by the degree of $Y$, which in turn is bounded by the maximum $d_0$ of the (normalized) leading coefficients of the finitely many elements of $\mathcal{E}$.

By Gotzmann's Regularity Theorem, there exists an integer $\rho=\rho(\mathcal{E})$ such that the Castelnuovo-Mumford regularity of $Y$ and every $D_j$ is $\leq \rho$ (actually the existence of such $\rho$ is older, presumably due to Mumford, but Gotzmann gave an explicit upper bound on $\rho$). Thus for the ideal sheaf $\mathcal{I}_Y$ of $Y$ in $\mathbb{P}^n$, $\mathcal{I}_{Y}(\rho)$ is globally generated, and $H^0(\mathbb{P}^n_F,\mathcal{O}(\rho))\to H^0(Y_j,\mathcal{O}_{Y}(\rho))$ is surjective, and the same is true for $D_j$. In particular, for distinct components $D_j$ and $D_k$ of $D$, the image of the restriction map $$H^0(\mathbb{P}^n_F,\mathcal{I}_{D_j}(\rho))\to H^0(D_k,\mathcal{I}_{D_j}\cdot\mathcal{O}_{D_k}(\rho))$$ has nonzero image. This can be checked after base change to an algebraic closure of $F$, and it follows there from the fact that $\mathcal{I}_{D_j}(\rho)$ is globally generated. Thus, there exists $f_{j,k}\in H^0(\mathbb{P}^n_F,\mathcal{O}(\rho)$ such that $f_{j,k}$ vanishes on $D_j$ but not on $D_k$. The element $f_k = \prod_{j\neq k} f_{j,k}$ is an element of $H^0(\mathbb{P}^n_F,\mathcal{O}((\ell-1)\rho))$ that vanishes on every $D_j$ with $j\neq k$, yet does not vanish on $D_k$. Thus, $f=\sum_k f_k$ is an element of $H^0(\mathbb{P}^n_F,\mathcal{O}((\ell - 1)\rho))$ that does not vanish identically on any $D_k$. The same is true for the power $f^s$. Thus, there exists a section of $H^0(\mathbb{P}^n_k,\mathcal{O}((d_0-1)!\rho)))$ whose zero scheme $Z$ on $Y$ does not contain any $D_j$. In particular, $Z$ is not contained in $\mathbb{P}^{n-1}_F$ (there may have been a faster way to arrange that).

By the same quasi-compactness arguments as above, the Hilbert polynomials of $Z$'s are bounded and have degree $<m$. Thus, letting $\mathcal{F}$ be the union of $\mathcal{E}_k$ for $k<m$ and the Hilbert polynomials of $Z$s as above, $\mathcal{F}$ is a finite set of Hilbert polynomials with smaller $m$. Thus, by the induction hypothesis, there exists an integer $d$ such that $Z(F')\setminus \mathbb{P}^{n-1}(F')$ is nonempty for some $F'/F$ of degree $<d$. Thus, this same $d$ works for $\mathcal{E}$, and the claim is proved by induction on $m$.