Prove that every nonzero prime ideal is maximal in $\mathbb{Z}[\sqrt{d}]$

Sounds good. In order to show that $I \cap \Bbb Z$ is a non-zero ideal, it is enough to notice that for $a+b\sqrt{d}\in I$ you have $(a+b\sqrt{d})(a-b\sqrt{d}) = a^2 -db^2 =:n\in \Bbb Z \cap I$. Now by writing $R = \Bbb Z[X]/(X^2-d)$ you can show that $R/P$ is a quotient of the finite ring $(\Bbb Z / n\Bbb Z)[X]/(X^2-d)$ where $X^2-d$ denotes the reduced polynomial in $(\Bbb Z/n \Bbb Z)[X]$. So indeed, $R/P$ is finite without zero divisors.

If the ideal is prime, almost by definition the quotient has no zero divisors.

On the other hand, since $R$ is a finite generated abelian group, the quotient $R/P$ is also a finitely generated abelian group, and to show it is finite it is enough to show that $R/P$ has finite exponent.

If $P$ is non-zero, there is a non-zero element $x=a+b\sqrt d$ in $P$, and then $e=a^2-db^2=(a-b\sqrt d)x\in P$; you can check easily that $e\neq0$. It follows that the product of every element of $R/P$ by $e$ is zero, and therefore the exponent of the abelian group $R/P$ divides $e$.