As a follow-up to this question: $\mathbb{Z}[\omega]$ is not a UFD, I am now trying to show that, for all $p$ such that $\mathbb{Z}[\omega]$ is a UFD (for example all primes less than $23$), that $x+y\omega$ has the form $u\alpha^{p}$ where $\alpha\in\mathbb{Z}[\omega]$ and $u$ is a unit in $\mathbb{Z}[\omega]$.
Disclaimer: Of course this is the case where $p\nmid x,y,z$.
My attempt
$$z^{p}=x^{p}+y^{p}=\prod_{j=0}^{p-1}(x+y\omega^{j}).$$ For $0\leq j<k\leq p-1$, any common ideal factor $J$ of $x+y\omega_{p}^{j}$ and $x+y\omega^{k}$ is a factor of $$(x+\omega^{j})-(x+y\omega^{k})=y\omega^{j}(1-\omega^{k-j})=vy(1-\omega)$$ for some $v\in\mathbb{Z}[\omega]^{\times}$. Since $y(1-\omega)| yp$, $J| (yp)$. Also, $J|z^{p}$, but $\gcd(yp,z^{p})=1$, so $J$ is the unit ideal. Hence the ideals $(x+\omega^{j})$ are coprime.
The product of these ideals if the $p^{\text{th}}$ power $(z)^{p}$, so unique factorisation implies that each factor is a $p^{\text{th}}$ power. Letting $j=1$ gives $(x+y\omega)=I^{p}$ for some ideal $I$. Therefore $I^{p}$ is trivial in the class group of $\mathbb{Q}(\omega)$. Since $p$ is regular, $I$ is trivial in the class group of $\mathbb{Q}(\omega)$, so $I$ is principal, say $I=(\alpha)$ with $\alpha\in\mathbb{Z}[\omega]$. Then $x+y\omega=u\alpha^{p}$ for some $\alpha\in\mathbb{Z}[\omega]$ and $u\in\mathbb{Z}[\omega]^{\times}$.
Can somebody tell me if this is correct, or if I can make it shorter?
