Just to see if I got it right.
$\Bbb Z [ \sqrt{-5}]$ has only 1 and -1 as a unit as seen here:
units of $\mathbb Z[\sqrt{-5}]$
I konow that. Let $a, b \in D$, $a$ and $b$ are said to be associated if $a\mid b$ and $b\mid a$. But as the only units are $1$ and $-1$, then there is no $ \neq b$ such that $a = ub$, as seen here:
Prove that $2$, $3$, $1+ \sqrt{-5}$, and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$.
So $2 + \sqrt{ -5} $ has no associates.
It sounds like you have the right idea, but you have omitted a lot of details. Let me fill them in.
Let $a$ be an associate of $2 + \sqrt{-5}$ in $\Bbb{Z}[\sqrt{-5}]$. Then we find that there exist $b,c\in\Bbb{Z}[\sqrt{-5}]$ such that $(2 + \sqrt{-5})b = a$ and $ac = 2 + \sqrt{-5}$. Combining these equations and rearranging, we find that $$ (2 + \sqrt{-5})(bc - 1) = 0. $$
Now, $2 + \sqrt{-5}$ is not a zero divisor in $\Bbb{Z}[\sqrt{-5}]$, so that we must have $bc = 1.$ Thus, $b$ and $c$ are units, so that $b,c\in\{1,-1\}.$ If $bc = 1,$ we must have $b = c,$ so that the only associates of $2 + \sqrt{-5}$ in $\Bbb{Z}[\sqrt{-5}]$ are $\pm(2 + \sqrt{-5}).$
