I wonder if there is a way to factor a number in $\mathbb{Z[\omega]}$ more "general"?, I currently use $(a+b\omega)(a+b\omega^2) = a^2 - ab + b^2 = \frac{(2a-b)^2 + 3b^2}{4}$ to find for example:
$13=(3+4\omega)(3+4{\omega}^2)$
but I have some issues with other numbers like $19$.
A good way to factor reducible natural primes $p$ (meaning $3$ and primes one greater than a multiple of $3$) is to first seek the unique form
$p=m^2+3n^2; m, n\in\mathbb Z_{\ge0}$
Then we render
$p=(m+n\sqrt{-3})(m-n\sqrt{-3})=[(m+n)+2n\omega][(m-n)-2n\omega]$
where the second link in the equality chain comes from rendering $\sqrt{-3}=2\omega+1$.
The two factors are of course conjugates of each other, so the second factor may be rendered as the conjugate of the first or vice versa. This gives two additional factorizations
$p=[(m+n)+2n\omega][(m+n)+2n\omega^2]=[(m-n)-2n\omega^2][(m-n)-2n\omega]$
Note that the coefficient on $\omega$ or $\omega^2$ will always be even, while (for prime or more generally odd inputs) the real terms will be odd.
Thus for instance $19=4^2+(3×1^2)$ gives
$19=(4+\sqrt{-3})(4-\sqrt{-3})=(5+2\omega)(3-2\omega)$
$=(5+2\omega)(5+2\omega^2)=(3-2\omega^2)(3-2\omega).$
