Is $\mathbb{Z}[x]/(x^2+1)$ isomorphic to $\mathbb{Z}[i]$?
You are certainly on the right track. Let's define a homomorphism $\phi:\mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$ as follows:
$$\phi(f(x)) = f(i)$$
Is $\phi$ a surjective homomorphism? Certainly: consider any $a+bi \in \mathbb{Z}[i]$. Note that $(bx + a) \mapsto (a+bi)$. Next, you should convince yourself that $\ker(\phi) = \langle x^2+1 \rangle$. At this point, we simply apply the isomorphism theorem:
$$\mathbb{Z}[x]/\langle x^2+1 \rangle \cong \mathbb{Z}[i]$$