Let $R$ be a finitely generated integral domain (over $\mathbb Z$) and let $I$ be a maximal ideal of $R$. We show $R/I$ is a finite field.  Let $K$ be an algebraic closure of $R/I$. Let $p$ be the characteristic of $K$.  Suppose $n$ elements generate $R$.  Then we can write $R/I= \mathbb Z[x_1,\ldots x_n]/(f_1,\ldots, f_m)$. Therefore, the first order sentence $\phi=\exists y_1,\ldots, y_n[f_1(y_1,\ldots,y_n)=0\wedge\cdots \wedge f_m(y_1,\ldots,y_n)=0]$ is true in $K$. There are two cases.

  If $p>0$, then since the first order theory of algebraically closed fields of characteristic $p$ is complete we have $\overline {\mathbb F_p}\models \phi$. It follows that $R/I$ embeds in $\overline {\mathbb F_p}$ and hence is finite being finitely generated. 

Next suppose $p=0$. By completeness of the theory of an algebraically closed field of characteristic $0$ models $\phi$.  It is a standard consequence of the compactness theorem of first order logic that there is an algebraically closed field of prime characteristic that models $\phi$. The previous paragraph now shows $R/I$ is a finite field.