Show that the norm of Gaussian prime of form $a + bi$ is always prime. I want to prove the partial converse of the following theorem:

Let $z \in \mathbb{Z}[i]$. If $N(z)$ - prime, then $z$ is also prime in $\mathbb{Z}[i]$.

I state the converse like this:

Let $z \in \mathbb{Z}[i]$, $z = a + bi$, $a,b \in \mathbb{Z}$ and both non-zero. If $z$ - prime in $\mathbb{Z}[i]$, then $N(z)$ is also prime.

I have tried to assume that $N(z) = a^2 + b^2 = n$ - composite, and then reach a contradiction, but unfourtunately had no success with it. Since $z$ - prime, writing $z = xy$ involves that either $x$ or $y$ is a unit. By multiplicativity of the norm in $\mathbb{Z}[i]$ I have $N(z) = N(x)N(y)$, and though $N(z) = N(x)$ or $N(z) = N(y)$. But from this point I don't know how to proceed and need some hint.
Would appreciate any help, thank you in advance.
 A: $z=a+ib$, so $\overline{z}=a-ib$, thus $z\overline{z}=a^2+b^2=N(z)$, showing that $z \mid N(z)$. $z$ is not a unit, so $N(z)$ is not $1$ and we have a prime factorization $N(z)=\prod_{i=1}^n p_i$. We have $z \mid N(z) =\prod_{i=1}^n p_i$. Thus for some $j$, we will have $z \mid p_j$. Set $p=p_j$. We have $zy=p$ for some $y \in \Bbb Z[i]$. Applying conjugates we get $\overline{z} \overline{y} =p$. This shows that both $z$ and $\overline{z}$ divide $p$. As $z$ does not divide $\overline{z}$ (nota bene: this step uses $a,b \neq 0$) and $z$ is prime, we have $\gcd(z,\overline{z})=1$, so the product $z\overline{z}$ must divide $p$ as well. But $z\overline{z}=N(z)$ is an integer, so from $N(z) \mid p$, we get $N(z)=p$ a required.
A: Based on the Lukas Heger answer and slightly modifying it I have the following proof.
We have $z=a+bi$ and it's conjugate $\overline{z}=a - bi$. Then $N(z) = a^2 + b^2 = (a+bi)(a-bi)=z \overline{z}$, thus $z \mid N(z)$. Since $z$ is prime in $\mathbb{Z}[i]$ and not a unit, then $N(z) \neq 1$ and we can factor it in primes as $N(z)=\prod_{i=1}^n p_i$, and so $z \mid p_j$ for some $j$. Let $p = p_j$, then $z \mid p$ implies that $\exists{y} \in \mathbb{Z}[i]$ such that $zy=p$. Since $p \in \mathbb{Z}$, we have $N(p) = p^2 = N(z)N(y)$. Since $z$ is not a unit, then $N(z) \neq 1$. And if $y$ is a unit, then it would contradict with the fact that for $z = a + bi$ we have both $a,b \neq 0$. So neither $z$ or $y$ are units, and so $N(z)=p$, as required.
