Let me add a self-contained answer which is **completely elementary** and avoids both the Nullstellensatz and Noether's normalization. It is a polished version of this article (already linked in Daniel Litt's comments). It partially overlaps with the answers of Daniel Litt and Guillermo Mantilla.

**Lemma 1.** Let $R$ be a UFD with infinitely many primes. Then an algebraic field extension $L$ of $F:=\operatorname{Frac}(R)$ cannot be finitely generated as an $R$-algebra.

**Proof.** Assume $L=R[y_1,\dots,y_n]$, with each $y_j$ being a root of a certain monic polynomial $p_j\in F[x]$. Taking the common denominator $d\in R$ of the coefficients of these polynomials $p_j$, we get that $y_j$ is **integral** over $R':=R[1/d]\subseteq F$. But then, given a prime $p\nmid d$ (here we use that $R$ has infinitely many primes), the same holds for $1/p\in L=R'[y_1,\dots,y_n]$. However, since $R'$ is still a UFD, this implies $1/p\in R'$ (as a UFD is integrally closed), contradiction. $\blacksquare$

**Lemma 2.** Given a field extension $K\subseteq L$, if $L$ is finitely generated as a $K$-algebra then the extension is algebraic, and in particular finite (meaning $\operatorname{dim}_K L<\infty$).

**Proof.** Assume $L=K[z_1,\dots,z_m]$. Since $L=K(z_1)[z_2,\dots,z_m]$, by induction $L$ is algebraic over $K(z_1)$. If $z_1$ is transcendental over $K$, then $R:=K[z_1]\cong K[x]$ satisfies the hypotheses of Lemma 1, which contradicts that $L$ is finitely generated as an $R$-algebra. So $z_1$ is algebraic over $K$, hence also $L$. (Note that the base case $m=1$ is obvious, as $1/z_1\in K[z_1]$ implies that $z_1$ is the root of some polynomial over $K$.) $\blacksquare$

**Theorem.** If $L$ is a field which is finitely generated, then $L$ is a finite field.

**Proof.** $L$ is (isomorphic to) the quotient of the ring $\mathbb{Z}[x_1,\dots,x_n]$ by a maximal ideal $M$. Observe that $M\cap\mathbb Z$ is a prime ideal of $\mathbb Z$. If $M\cap\mathbb Z=\{0\}$, then $\mathbb Z$ embeds into $L$; but this contradicts Lemma 1 (with $R:=\mathbb Z$)! Hence $\mathbb F_p$ embeds into $L$ for some prime number $p$ and Lemma 2 gives that $L$ is finite. $\blacksquare$

Bourbaki, N. Algèbre Commutative. Chapitre V. Entiers, Hermann, Paris, 1964. (p. 68, cor. 1)$\endgroup$ – Mike Pierce May 19 '18 at 6:39