Fields finitely generated as $\mathbb Z$-algebras are finite?

Yes!
Consider the morphism $f:\mathbb Z\to k$ and the ideal $\mathfrak m=f^{-1}(0)\subset \mathbb Z$.
Since $\mathbb Z$ is a Jacobson ring and $(0)\subset k$ is maximal, $\mathfrak m$ is maximal too and we obtain a morphism $\bar f:\mathbb F_p\to k$.
Since $k$ is finitely generated over $\mathbb F_p$ and is a field, it is actually a finite extension ("Zariski's version of the Nullstellensatz") and thus $k$ is (set-theoretically!) finite.

Edit
Considering the comments below , I had better state explicitly the theorem I have used.
Theorem
Let $f:A\to B$ be a finitely generated $A$-algebra.
If $A$ is a Jacobson ring, , then $B$ is also Jacobson and for every maximal ideal $\mathfrak m\subset B$ the ideal $f^{-1}(\mathfrak m)\subset A$ is also maximal.

New Edit
Here is a proof in the style of algebraic geometry, due to Akaki Tikaradze.
If $k$ has characteristic $p$ we conclude as above, using Zariski.
If $\operatorname{char}k=0$, the morphism $f:\mathbb Z\to k \:$ is injective, so $A$ has no $\mathbb Z$-torsion and is thus $\mathbb Z$-flat.
But then $\operatorname{Spec}(k)\to\operatorname{Spec}(\mathbb Z)$ is open (flat+finite presentation $\implies$ open), i.e. $(0)\in\operatorname{Spec}(\mathbb Z)$ is open. Contradiction.
(This proof is absurdly sophisticated but it will probably appeal to scheme-theory addicts. Take it as some kind of joke...)

In case you are interested, here is a proof which also uses Zariski's lemma but no difficult theorems about Jacobson rings.

Write $R$ for the image of the unique ring homomorphism $\mathbb{Z} \to k$, so that $k$ is a finitely generated $R$-algebra and hence a finite extension of the fraction field of $R$ by Zariski's lemma. Thus it suffices to show that $R = \mathbb{F}_p$, which is to say $k$ has positive characteristic. If $R = \mathbb{Z}$, meaning $k$ has characteristic zero, then $k$ is a number field which is a finitely generated ring. But this is impossible: if we write $k = \mathbb{Z}[\alpha_1,\dots,\alpha_r]$, then one can choose $n \in \mathbb{Z}$ so that all the denominators of coefficients in the minimal polynomials over $\mathbb{Q}$ of $\alpha_1,\dots,\alpha_r$ divide $n$. This implies that $k$ is integral over $\mathbb{Z}[1/n]$. Then $\mathbb{Z}[1/n]$ must be a field, which is absurd.

This is a mild generalization of Justin Campbell's answer, and the proof is almost the same as that of Proposition 7.8 in Atiyah and MacDonald's Introduction to Commutative Algebra. (This proposition is sometimes called the Artin-Tate Lemma.) The notation has been designed to make the analogy as clear as possible.

Let $A\subset B\subset C$ be a tower of domains such that $B$ is the field of fractions of $A$, and $C$ is a field which is finitely generated as an $A$-algebra. Then (a) $C$ is a finite extension of $B$, and (b) $B$ is a finitely generated $A$-algebra.

As (a) results immediately from Zariski's Lemma, it suffices to prove (b). The following statement is ironically stronger and easier to prove than (b):

Assume that $C=A[x_1,\dots,x_m]$; that $y_1,\dots,y_n$ is a $B$-basis of $C$ with $y_1=1$; and that we have $$x_i=\sum_jb_{ij}y_j,\quad y_iy_j=\sum_kb_{ijk}y_k$$ with $b_{ij}$, $b_{ijk}$ in $B$. Then the $A$-algebra $B$ is generated by the $b_{ij}$ and the $b_{ijk}$.

The proof is straightforward, and left to the reader.