I asked a question before. I am asking the same question again but in different manner with less confusion.
How can we prove that a Gaussian integer is prime in $\mathbb{Z}[i]$ only when the norm of $z$, $N(z)$ is a prime number in $\mathbb{Z}$? Is there any Gaussian integer $z$ ( not on real or imaginary axes) whose norm is composite in $\mathbb{Z}$ but $z$ is itself a prime in $\mathbb{Z}[i]$?
On
There is a notion of prime in the ring $\mathbb{Z}[x]$ as a prime is defined to an irreducible element in an integral domain, and $\mathbb{Z}[x]$ is an integral domain, and the elements in the gaussian integers which have a prime norm are not always irreducible, if they are of form $4k+1$, then they are reducible(i mentioned in the answer of your previous question). But if the norm is not of form $4k+1$ and it is also a prime norm, then it is of couse irreducible, and hence can be called as a prime in $\mathbb{Z}[x]$
The comment by Wojowu already supplies the full answer.
So you get a counterexample by considering Gaussian integers like $\pm 3$ and $\pm 3i$. They are Gaussian primes, but their norms, $N(\pm 3) = N(\pm 3i) = 9$ are composite (9 is a composite number).
As is also contained in Wojowu's comment, the only counterexamples are Gaussian primes of the type $\pm p$ and $\pm ip$ where $p$ is a positive prime in $\mathbb{Z}$ such that $p\equiv 3 \pmod 4$ (i.e. $p$ leaves a remainder of $3$ when divided by $4$). Their norms $p^2$ are composite.