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Oh, I think the answer is definitely yes!


Let $\{k \to x,y\}$ be any solution of $x^2 - (k^2+1)y^2 = k^2$, and let $K$ be the set of $k$ for which a solution has $0 < k < y-1$. In a paper recently submitted to Glasnik Matematicki we call these solutions exceptional solutions. Andrej's conjecture is that for any $k$ there is at most 1 exceptional solution.

One interesting result we obtain is that, if $k \in K$, then $y < (2 - \sqrt{3})k$.

A particular feature of this Pell equation is its symmetry wrt $k$ and $y$. These are interchangeable, so for any solution $\{k \to x,y\}$ there is a corresponding solution $\{y \to x,k\}$.

It follows that if $k \neq y \pm{1}$, then either $k \in K$ or $y \in K$.

Now, for any $k \geq 2$, we have 3 solution classes $(x_n, y_n)$ with $y_0 = \{0, k-1, -(k-1)\}$. For any $n > \{0, 0, 1\}$ we have $y_n > k-1$ and so $\{k \to x_n,y_n\}$ is exceptional, ie. $y_n \in K$.