Oh, I think the answer is definitely yes!
Let $\{k \to x,y\}$ be any solution of (), and let $K$ be the set of $k$ for
which () has any solution with $0<k<y-1$. In a paper recently recently
submitted to Glasnik Matematicki we call these solutions
exceptional solutions to ().
One interesting result we obtain is that, if $k \in K$, then
$y < (2 - \sqrt{3}k$.
A particular feature of () is its symmetry wrt $k$ and $y$. 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 $y \in K$.
Jim White
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