Infinite solutions of Pell's equation $x^{2} - dy^{2} = 1$ 
Let $d > 1$ be a squarefree integer. Prove that the equation $x^{2} - dy^{2} = 1$ has infinitely many solutions in $\mathbb{Z} \times \mathbb{Z}$.

What I have done: let $ \ \mathbb{K} = \mathbb{Q}[\sqrt{d}]$. If $ d \not\equiv 1$ mod $4$, then $O_{\mathbb{K}} = \mathbb{Z}[\sqrt{d}]$ and the statement follows by the Dirichlet Unit Theorem. What about the case $ d \equiv 1 $ mod $4$ ?

You can handle the case $d\equiv 1 \pmod{4}$ in essentially the same way. For any unit $u$, $u^6=a+b\sqrt{d}$ with $a,b\in\mathbb Z$ and $a^2-db^2=1$, then infinitude follows from Dirichlet's Unit Theorem.
If $u$ is a unit then $u^2=\frac{\alpha}{2}+\frac{\beta}{2}\sqrt{d}$ with $\alpha,\beta\in \mathbb Z$, $\alpha \equiv \beta \pmod{2}$ (by definition of $O_{\mathbb K}$) and
$$
N(u^2)=\frac{\alpha^2}{4}-\frac{d\beta^2}{4}=1
$$
Then either:
Case 1 $\alpha\equiv\beta\equiv 0 \pmod{2}$. Then $\frac{\alpha}{2},\frac{\beta}{2}\in\mathbb Z$ and it follows from multiplication in $\mathbb K$ that $\exists a,b\in \mathbb Z$ with $u^6 =a+b\sqrt{d}$.
Case 2 $\alpha\equiv\beta\equiv 1 \pmod{2}$. Since $\alpha^2-d\beta^2=4$ and $\alpha^2\equiv\beta^2\equiv 1 \pmod{8}$ this case cannot arise unless $d\equiv 5\pmod{8}$. Then
$$
u^6 = \frac{1}{8}\left(\alpha(\alpha^2+3d\beta^2)+\beta(3\alpha^2+d\beta^2)\sqrt{d}\right) \\
\alpha^2+3d\beta^2\equiv 3\alpha^2+d\beta^2\equiv 0 \pmod{8} \\
a=\frac{\alpha(\alpha^2+3d\beta^2)}{8},b=\frac{\beta(3\alpha^2+d\beta^2)}{8}
$$
and $a,b\in\mathbb Z$.
