the Gaussian integers are isomorphic to $\mathbb{Z}[x]/(x^2+1)$ I am trying to prove that $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$.

My initial plan was to use the first isomorphism theorem.  I showed that there is a map $\phi: \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$, given by $\phi(f)=f(i).$  This map is onto and homorphic.  The part I have a question on is showing that the $ker(\phi) = (x^{2}+1)$. 

One containment is trivial, $(x^2+1)\subset ker(\phi)$.  To show $ker(\phi)\subset (x^2+1)$, let $f \in ker(\phi)$, then f has either $i$ or $-i$ as a root.  Sot $f=g(x-i)(x+i)=g(x^2+1).$ 
How can I prove that $f \in \mathbb{Z}[x]\rightarrow g \in \mathbb{Z}[x]$?
Let me elaborate on Daniel Fischers comment. You have a ring homomorphism $\phi: \mathbb Z[x]\to\mathbb Z[i]$ given by $x\mapsto i$. Take $f \in \ker \phi$. By the division algorithm,
$$ f = q\cdot(x^2+1) + r,$$
where $\deg r &lt; 2$ and $q,r\in\mathbb Z[x]$, since $x^2+1$ is monic. Applying $\phi$ to this equation yields
$$ 0 = \underbrace{\phi(q)\cdot(i^2+1)}_0 + \phi(r).$$
Since $r$ is of degree $&lt;2$, we can write $r = ax+b$ for some $a,b\in\mathbb Z$. Then $\phi(r)=0$ gives
$$ai+b=0.$$
This equation in $\mathbb Z[i]$ implies $a=b=0$, so we have $r=ax+b=0\in\mathbb Z[x]$ and therefore
$$
f = q\cdot (x^2+1) \in \langle x^2+1\rangle.
$$
We conclude that $\ker \phi \subseteq \langle x^2+1\rangle$.
You can define your isomorphism as follows

$\phi:\mathbb{Z}[i]\rightarrow \mathbb{Z}[x]/\langle x^2+1 \rangle$ 

$\phi(a+bi)=(a+bx)+\langle x^2+1 \rangle$ 

If you compute $[a+bx+\langle x^2+1 \rangle]\times [c+dx+\langle x^2+1 \rangle]$

We get $ac+(ad+bc)x+bdx^2+\langle x^2+1 \rangle$. We reduce by long division to get, 

$(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.

Therefore,

$\phi([a+bi][c+di])=(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.

Which is exactly what you want.

The map is obviously bijective and a ring homomorphism.