The problem is: Show that if $p(\neq 3)$ is a prime in $\mathbb Z$ and $p \neq a^2+b^2-ab$ for any $a,b \in \mathbb Z$, then $p$ is a prime element in the ring $R = \mathbb Z[\omega]$.
My approach: I was trying by contradiction that if $p$ isn't prime in $R$, then there exists $(a+b\omega)$ and $(c+d\omega)$ such that $p\mid(ac-bd)$ and $p\mid(ad+bc-bd)$ but $p\nmid(a+b\omega)$ and $p\nmid(c+d\omega)$; but then how to use $p$ is prime in $\mathbb Z$?
Am I approaching correctly?
A small hint is warmly appreciated.
If $x|p$, where $x \in \mathbb{Z}[\omega]$, then $x \bar{x} | p \bar{p} = p^2$. Since $x \bar{x} \in \mathbb{Z}_+$, you have only three options for $x \bar{x}$. Try to rule out these three posibilities (for example, when $x \bar{x}=1$, $x$ is a unit, so our alleged factorisation wasn't quite proper).