= Solution
The characteristic of $K$ is either a prime $p$ or zero. In characteristic $p$, it is a <finitely generated algebra> over $\mathbb F_p$. The course's <Zariski lemma> says that a <field> finitely generated as an algebra over a <field> is a finite algebraic extension of that <field>. Therefore
$$
\boxed{\operatorname{char}K=p\Longrightarrow |K|=p^d\text{ for some finite }d}.
$$
It remains to exclude characteristic zero. Write $K=\mathbb Z[a_1,\ldots,a_m]$. Because $K$ is then a <field> containing $\mathbb Q$, also $K=\mathbb Q[a_1,\ldots,a_m]$. <Zariski lemma> makes $K/\mathbb Q$ finite algebraic. Each $a_i$ has a monic equation with rational coefficients. Choose a positive integer $N$ clearing all their coefficient denominators. Each $a_i$ is then integral over $A=\mathbb Z[1/N]$, so $A[a_1,\ldots,a_m]$ is integral over $A$. This algebra equals $K$: it contains the given integer algebra $K$, and $1/N$ already lies in the <field> $K$.
An <integral field extension forces the base domain to be a field>. To see it here, for a nonzero $a\in A$ its inverse in $K$ satisfies
$$
(a^{-1})^r+c_{r-1}(a^{-1})^{r-1}+\cdots+c_0=0,\qquad c_i\in A.
$$
Multiplying by $a^{r-1}$ gives
$$
a^{-1}=-(c_{r-1}+c_{r-2}a+\cdots+c_0a^{r-1})\in A.
$$
Thus $A$ would be a <field>. But choose a prime $\ell\nmid N$. Reduction modulo $\ell$ is a well-defined map $\mathbb Z[1/N]\to\mathbb F_\ell$, so the nonzero element $\ell$ is not invertible in $A$. This contradiction rules out characteristic zero. Therefore the <finite-field theorem for finitely generated integer algebras> yields
$$
\boxed{K\text{ is a finite field}}.
$$
This uses <Zariski lemma> and elementary integrality, not a claim that a <field> merely finitely generated as a <field extension> must be algebraic.
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