$a+b\sqrt{-3}$ and $a-b\sqrt{-3}$ are coprime in $\mathbb{Z}+ \omega \mathbb{Z}$

Actually, there's no contradiction.

Even if $a,b$ are relatively prime, the claim is false.

For example, let

$$y = 1 + \sqrt{-3}$$

$$z = 1 - \sqrt{-3}$$

Then $y,z$ are non-units, but each divides the other, hence they have a common non-unit factor.

So many things that need clarification here. What are $a$ and $b$? Are they integers, that is $a, b \in \mathbb Z$? And are you sure you're only looking at numbers of the form $a + b \sqrt{-3}$, and not also numbers of the form $a + b \omega$?

It needs to be said explicitly that $$\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2}.$$ With that in mind, we can rewrite $1 - \sqrt{-3}$ and $1 + \sqrt{-3}$ as $-2 \omega$ and $2 + 2 \omega$ respectively. We have changed $a$ and $b$, of course, but it should still be clear that $2 \omega$ and $-2 \omega$ have at least one prime factor in common, as do $2 - 2 \omega$ and $2 + 2 \omega$.

Also note that $N(a + b \omega) = a^2 - ab + b^2$.