A finitely generated $\mathbb{Z}$-algebra that is a field has to be finite

To prove Nullstellensatz over $\mathbb{Z}$: as the morphism $f: \mathrm{Spec}(R)\to\mathrm{Spec}(\mathbb Z)$ is of finite type, a theorem of Chevalley says that the image of any constructible subset is constructible. So the image of any closed point by $f$ is a point which is a constructible subset. This can not be the generic point of $\mathrm{Spec}(\mathbb Z)$, so it must be a closed point.

Note that this does not hold in general. For example, over the ring of $p$-adic integers, the ideal $(pX-1)\mathbb{Z}_p[X]$ is maximal, but its preimage in $\mathbb{Z}_p$ is $0$ and it not maximal.

[EDIT] Another proof using Noether's normalization lemma: Noether's normalization lemma over a ring A: if a maximal ideal $\mathfrak m$ of $R$ is such that $\mathfrak m\cap \mathbb Z=0$, then $R/\mathfrak m$ is finite type over (and contains) $\mathbb Z$. So there exits $f\in\mathbb Z$ non-zero and a finite injective homomorphism $\mathbb Z_f[X_1,\dots, X_d]\hookrightarrow R/\mathfrak m$. But then $\mathbb Z_f[X_1,\dots, X_d]$ must be a field. This is impossible because the units of this ring are $\pm f^k$, $k$ relative integers.

Let $R$ be a finitely generated $\mathbb{Z}$-algebra, and $\mathfrak{m}\subset R$ are maximal ideal. We wish to show $R/\mathfrak{m}$ is a finite field.

Let $i: \mathbb{Z}\to R$ be the unique ring map; then $i^{-1}(\mathfrak{m})$ is a maximal ideal in $\mathbb{Z}$ (as $R$ is finitely generated over $\mathbb{Z})$, and thus $\mathbb{Z}/i^{-1}(\mathfrak{m})$ is a finite field $\mathbb{F}_p$ for some prime $p$. As $R$ is finitely generated over $\mathbb{Z}$, $R/\mathfrak{m}$ is finitely generated over $\mathbb{F}_p$. But all finite field extensions of $\mathbb{F}_p$ are still finite, completing the proof.

This is not an answer to your question, but let me point out that the Ax-Grothendieck theorem is now easy to prove using E-polynomials (Hodge-Deligne polynomials). If $f:X \to X $ is an injective endomorphism of a complex algebraic variety, then $E(X) = E(f(X))=E(X)-E(X\setminus f(X))$. So $E(X\setminus f(X))=0$ and $X\setminus f(X) = \emptyset$, because the degree of a constructible set is twice its dimension. Since one supposes the mixed Hodge theory, this proof is not trivial at all. But at least for me, this looks more natural.