I have seen this as an answer to Prove that $p$ is prime in $\mathbb{Z}[\sqrt{-3}]$ if and only if $x^2+3$ is irreducible in $\mathbb{F}_p[x]$.
If you assume that $x^2+3$ is irreducible, then you have that $$\mathbb{Z}[\sqrt{-3}]/(p)\cong \mathbb{Z}[x]/(p,x^2+3)\cong \mathbb{F}_p[x]/(x^2+3)$$
However, I am having a very difficult time proving this.
$\textbf{Attempt:}$
Suppose $x^2+3$ is irreducible in $\mathbb{Z}_p[x]$. Then we have that $\frac{\mathbb{Z}_p[x]}{(x^2+3)}$ is a field. Let $\phi: \mathbb{Z}[x] \to \frac{\mathbb{Z}_p[x]}{(x^2+3)}$ be a homomorphism such that $p(x) \to q(x) \text{mod} (x^2+3)$ where $q(x) \in \mathbb{Z}_p[x]$. Let $K=(p,x^2+3)=\{ph(x) + (x^2+3)r(x)\}$. Then $\phi(K)=0$, thus $K$ is the kernel. Then by the first isomorphism theorem we have that $\frac{\mathbb{Z}[x]}{(p,x^2+3)} \cong \frac{\mathbb{Z}_p[x]}{(x^2+3)}$.
I am sure there are some errors in my attempt, furthermore I am having a difficult time defining these ring homomorphism correctly. Also, I am completely stuck as to how to prove that $$\mathbb{Z}[\sqrt{-3}]/(p)\cong \mathbb{Z}[x]/(p,x^2+3)$$