 Solution

Let $K$ be a field finitely generated as a $\mathbb{Z}$-algebra. If $K$ has positive characteristic $p$, it is a finitely generated algebra over ${\mathbb{F}}_p$; Zariski lemma makes it a finite algebraic extension of the finite field ${\mathbb{F}}_p$, so $K$ is finite.

Suppose instead that $K$ has characteristic zero. Then $K$ is a finitely generated $\mathbb{Q}$-algebra and Zariski lemma makes it a number field. For algebra generators ${\alpha }_1,\dots ,{\alpha }_r$, choose a nonzero integer $N$ such that every ${\alpha }_i$ is integral over $\mathbb{Z}[1/N]$. The whole algebra $K=\mathbb{Z}[{\alpha }_1,\dots ,{\alpha }_r]$ would then be integral over $\mathbb{Z}[1/N]$. But for a prime $q\nmid N$, the element $1/q\in K$ is not integral over the integrally closed domain $\mathbb{Z}[1/N]$, a contradiction. Hence every field finitely generated over the integers is finite, and in particular no infinite field has that property.