All the units are satisfied Pell's equation $a^2-2b^2=\pm1$ for $\mathbf{Z}[\sqrt{2}]$, $a,b\in\mathbf{Z}$.
Here is my proof:
Let $a+b\sqrt{2}$ be a unit $\in\mathbf{Z}[\sqrt{2}]$. This implies $$(a+b\sqrt{2})(c+d\sqrt{2})=1, c+d\sqrt{2}\in\mathbf{Z}\sqrt{2}$$ $$\implies \mathrm{norm}((a+b\sqrt{2})(c+d\sqrt{2}))=1$$ $$\implies \mathrm{norm}(a+b\sqrt{2})\cdot \mathrm{norm}(c+d\sqrt{2})=1$$ $$\implies (a+b\sqrt{2})(a-b\sqrt{2})(c+d\sqrt{2})(c-d\sqrt{2})=1$$ $$\implies (a^2-2b^2)(c^2-2d^2)=1$$ $$\implies \left[ a^2-2b^2=1\text{ and } {c}^2-2{d}^2=1\right]\text{ or } \left[ a^2-2{b}^2=-1\text{ and }{c}^2-2{d}^2=-1\right]$$ $$\implies {a}^2-2{b}^2=\pm1$$