Gaussian Integers and Quotient Rings Show that $\mathbb{Z}[i]/(2+3i)$ is a finite field.

By applying some isomorphisms and realizing $\mathbb{Z}[i]$ as the quotient ring of $\mathbb{Z}[x]/(x^{2}+1)$, I found that:
$\mathbb{Z}[i]/(2+3i) \cong \mathbb{Z}[x]/(x^2+1,2+3x)$
However, I run into trouble as I can't divide $x^{2}+1$ by $2+3x$ in $\mathbb{Z}[x]$. What would be another way to approach the problem?
Thanks for the help.
 A: Firstly, consider the norm $N:\mathbb{Z}[i]\rightarrow\mathbb{Z}[i]$ given by $a+bi\mapsto a^2+b^2$. Because $N(2+3i)=13=1(\mathrm{mod} 4)$ which is a prime element in $\mathbb{Z}$, $2+3i$ is a prime element in $\mathbb{Z}[i]$. Hence $\mathbb{Z}[i]/(2+3i)$ is an integral domain.
Secondly, $13\in(2+3i)$ and then $\mathbb{Z}[i]/(2+3i)\cong\frac{\mathbb{Z}[i]/(13)}{(2+3i)/(13)}$. Because $\mathbb{Z}[i]/(13)\cong\mathbb{F}_{13}[i]$ is finite, $\mathbb{Z}[i]/(2+3i)$ is also finite.
Therefore, $\mathbb{Z}[i]/(2+3i)$ is a finite integral domain, which is a finite field.
PS: Since $\mathbb{Z}[i]$ is a Dedekind domain, one obtains directly that any quotient ring of $\mathbb{Z}[i]$ is finite.
A: $2+3i$ is prime in $\mathbb Z[i]$ since $N(2+3i)=13$ is a prime in $\mathbb Z$. But $\mathbb Z[i]$ is a PID, so $\mathbb Z[i]/(2+3i)$ is a field.
Why is $\mathbb Z[i]/(2+3i)$ finite? Note that the elements of $\mathbb Z[i]/(2+3i)$ are the residues modulo $2+3i$, that is, the elements of $\mathbb Z[i]$ whose norm is $<13$. (Don't forget that $\mathbb Z[i]$ is Euclidean.)
