Finding the units in $\mathbb{Z}[\sqrt{-3}]$ using the norm

The norm is multiplicative: $N(\alpha\beta) = N(\alpha)N(\beta)$.

In particular, if $\alpha$ is a unit, then there exists $\beta$ such that $\alpha\beta=1$, so $1= N(1) = N(\alpha\beta) = N(\alpha)N(\beta)$. Thus, if $\alpha$ is a unit, then $N(\alpha)=\pm 1$.

Conversely, if $N(\alpha)=\pm 1$, then $\alpha\overline{\alpha}=1$ or $\alpha(-\overline{\alpha}) = 1$, so $\alpha$ is a unit.

(In fact, since the norm is always positive you can ignore one of the possibilities, but notice that I never used that we are working in $\mathbb{Z}[\sqrt{-3}]$. All we are using is that we have a multiplicative map $N\colon R\to\mathbb{Z}$ to get the necessity; and the sufficiency from the fact that we can evaluate this map $N$ via a unary operator, in this case the map $a+b\sqrt{-3}\mapsto a-b\sqrt{-3}$. So the argument easily generalizes to any $\mathbb{Z}[\sqrt{d}]$ with $d$ squarefree, and the ring of integers of any Galois extension of $\mathbb{Q}$)