Show that $\mathbb Q(\sqrt{5})$ contains infinitely many elements with a uniform bound on their partial quotients, by checking that the numbers $\overline{[1^{r+1},4,2,1^r,3]}$ for $r\ge 0$ all lie in $\mathbb Q(\sqrt 5)$ (here $1^r$ denotes the string $1,1,...,1$ of length $r$). Can you find a similar pattern in any real quadratic field $\mathbb Q(\sqrt d)$?
Since $u$ is a strictly periodic continued fraction, we know that $$ u=\frac{up_k+p_{k-1}}{uq_k+q_{k-1}}\quad\text{for}\quad u=\overline{[a_0;a_1,...,a_k]} $$ where $\frac{p_k}{q_k}=[a_0;a_1,...,a_k]$ and we have $$ u^2q_k+u(q_{k-1}-p_k)-p_{k-1}=0 .$$ $u$ has discriminant $\Delta=(q_{k-1}+p_k)^2-4(-1)^k$. So it suffices to check whether $\Delta=5l^2$, $l\in\mathbb Z$.
But how to show that for every $r$ this is true. I cannot come up with an induction.