Let $\mathcal{O}$ be the number ring of $\mathbb{Q}(\sqrt{-3})$, then $$\mathcal{O}=\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2} \right]=\mathbb{Z}[\omega]$$ I am told that $\mathcal{O}$ is a PID. I have factorized the ideal $7\mathcal{O}$ as follows: $$7\mathcal{O}=(7, \omega -5)(7, \omega -3)$$ However, as $\mathcal{O}$ is a PID I would like to express both of these ideals as principal ideals. I know from the factorisation of $7\mathcal{O}$ that both $(7, \omega -5)$ and $(7, \omega -3)$ must have norm $7$. Therefore I just have to look up for an element with norm $7$ (the norm of a principal ideal is the norm of its generator).
Let $\alpha=x+y\omega$, $x,y\in \mathbb{Z}$. Then $$N(\alpha)=x^2-y^2\omega^2=x^2-y^2\left(-\frac{1}{2}+\frac{\sqrt{-3}}{2} \right)=7$$ $$2x^2+y^2-y^2\sqrt{-3}=7$$ However, this doesn't have any solutions in $\mathbb{Z}$. Where am I going wrong?