I am working through Neukirch's Algebraic Number Theory on my own. Exercise 6 in Section 1 (page 5) is to show that the ring $\mathbb{Z}[\sqrt{d}]$, for any squarefree rational integer $d>1$, has infinitely many units.
I know that in $\mathbb{Z}[\sqrt 2]$ there are infinitely many units, because $(\sqrt{2} + 1)(\sqrt{2} - 1) = 1$ and then taking $n$th powers shows that $(\sqrt{2} + 1)^n$ is a unit for any $n\ge 1$.
Similarly in $\mathbb{Z}[\sqrt{3}]$, we have $(2+\sqrt{3})(2-\sqrt{3}) = 1$, and then $(2+\sqrt{3})^n$ for $n\ge 1$ is an infinite family of units.
I can find other "fundamental units" for other specific values of $d$.
But it seems I have to show that the (Pell) equation $a^2 - db^2 = \pm 1$, for any $d>1$, has an integer solution $(a, b) \ne (\pm 1, 0)$, because I know if I can find one solution, then I can get infinitely many. But from my limited knowledge of Pell's equation this is a difficult problem (using techniques such as continued fractions.)
Maybe there is a simpler nonconstructive proof that I'm missing. Any hints or suggestions?