What does the correspondence theorem tell us about the ideals of $\Bbb Z[x]$ that contain $x^2+1$?
If I define $\varphi : \Bbb Z[x] \to \Bbb Z[i]$ as $x \longmapsto i$, then $\ker \varphi = (x^2+1)$.
So by the correspondence theorem an ideal $I$ containing $(x^2+1)$ in $\Bbb Z[x]$ corresponds to an ideal $J = \varphi(I) \in \Bbb Z[i]$. However $\Bbb Z[i]$ is a PID so $J = (a+bi)$ for some $a,b \in \Bbb Z$.
Thus $$I = \varphi^{-1}(J) = \varphi^{-1}((a+bi)) = \{g(x) \in \Bbb Z[x] \mid \varphi(g(x)) = g(i) \in (a+bi) \}.$$
And $g(i) \in (a+bi)$ if and only if $g(i)$ is of the form $(c+di)(a+bi)$ for $c,d,a,b \in \Bbb Z$.
Is there some more information I could get out of here or is this all that the correspondence theorem tells us?
Because of the correspondence theorem, for $K$ an ideal of $\mathbb{Z}[x]$ such that $\langle x^2+1\rangle\subseteq K$, we have $\varphi^{-1}(\langle a+bi\rangle)=K\iff \langle a+bi\rangle=\varphi(K)$.
Can you verify that $\langle a+bi\rangle=\varphi(\langle x^2+1,a+bx\rangle)$?
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Something else to think about: when is $\langle a+bi\rangle=\langle c+di\rangle$?
As you know, $\mathbb{Z}[i]$ is a PID. The only units of $\mathbb{Z}[i]$ are $1,-1,i,-i$.
$\langle a+bi\rangle=\langle c+di\rangle\iff c+di\in\{a+bi,-a-bi,-b+ai,b-ai\}$.
This shows you when it is that $\langle x^2+1,a+bx\rangle=\langle x^2+1,c+dx\rangle$.