What are the units in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$?

Just a check that Pell unit provided by @Rene Schipperus is indeed independent of the other Pell units. For assume that we have $$(3+2\sqrt{2})^m(2+\sqrt{3})^n(5+2\sqrt{6})^p=1$$ Applying the Galois maps $\sqrt{2}\mapsto - \sqrt{2}$ ,$\sqrt{3}\mapsto - \sqrt{3}$ we get $$(3-2\sqrt{2})^m(2-\sqrt{3})^n(5+2\sqrt{6})^p=1$$ so $$(3+2\sqrt{2})^m(2+\sqrt{3})^n=(3-2\sqrt{2})^m(2-\sqrt{3})^n$$ and therefore $$\left(\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\right) ^m=\left(\frac{2-\sqrt{3}}{2+\sqrt{3}}\right)^n$$ so both must be in $\mathbb{Q}$, since $\mathbb{Q}(\sqrt{2})\cap \mathbb{Q}(\sqrt{3})=\mathbb{Q}$. Now, from $\left(\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\right) ^m \in \mathbb{Q}$, applying the Galois map $\sqrt{2}\mapsto -\sqrt{2}$ we conclude that $$\left(\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\right) ^m=\left(\frac{3-2\sqrt{2}}{3+2\sqrt{2}}\right) ^m$$ so the common value must be $\pm 1$. We conclude that $m=0$. Similarly we get $n=0$.

ADDED: Using WolframAlpha I got this result. The surprise was $$2+\sqrt{3}=\left(\frac{\sqrt{2}+\sqrt{6}}{2}\right )^2$$ Moreover, the units $$1+\sqrt{2}, \frac{\sqrt{2}+\sqrt{6}}{2}, \sqrt{2}+\sqrt{3}$$ form a fundamental system.

Note that $\frac{\sqrt{2}+\sqrt{6}}{2}=\frac{1+\sqrt{3}}{\sqrt{2}}$