Show that the powers of sqrt(2) + 1 are all distinct, and so (unlike Gaussian integers) there are infinitely many invertible elements in Z[sqrt(2)].
Suppose $\displaystyle (1+\sqrt2)^n=(1+\sqrt2)^k $ where $\displaystyle n>k $.
We then would have $\displaystyle (1+\sqrt2)^{n-k}=1 $ which is impossible since $\displaystyle 1+\sqrt2>1\implies (1+\sqrt2)^{n-k}>1 $
Now observe $\displaystyle (1+\sqrt2)^n\cdot (\sqrt2-1)^n = 1 $.