Showing that $\mathbb{Z}[i]/I$ is a finite field whenever $I$ is a prime ideal, and also finding its cardinality?

How to show that $\mathbb{Z}[i]/I$ is a finite field whenever $I$ is a prime ideal? Is it possible to find the cardinality of $\mathbb{Z}[i]/I$ as well?

I know how to show that it is an integral domain, because that follows very quickly.

Let $\alpha \in I$ be an element, with norm $N(\alpha)$. Any element $x \in \mathbb{Z}[i]$ can be written as $q \alpha + r$ with $N(r) < N(\alpha)$. So every element of $\mathbb{Z}[i]/I$ (viewed as an equivalence class) contains an element of norm smaller than $N(\alpha)$, and there are only finitely many such elements.

We don't have to use the assumption that $I$ is prime for the finiteness.

Another approach, especially if you’re only interested in counting $\Bbb Z[i]/I\,$: There’s a theorem in algebraic number theory that if $I$ is a nonzero principal ideal of the ring of integers $R$ in an algebraic number field $K$, say with $I=(\xi)$, then the cardinality of $R/I$ is $\big|\text N(\xi)\big|$, where N is the field-theoretic norm from $K$ down to $\Bbb Q$.

In the case $K=\Bbb Q(i)$, every $I$ is principal and the norm already is positive, so you can drop the absolute-value bars. In other words, if $I=(a+bi)$, then the cardinality of $\Bbb Z[i]/I$ is $a^2+b^2$. This has nothing to do with whether $I$ is prime.