Oh, I think the answer is definitely yes!<br><br>
<br>
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 <i>Glasnik Matematicki</i> we call these solutions
<i>exceptional</i> solutions to (*). 
<br><br>One interesting result we obtain is that, if $k \in K$, then
$y < (2 - \sqrt{3}k$.
<br><br>A particular feature of (*) is its symmetry wrt $k$ and $y$. So 
for any solution $\{k \to x,y\}$ there is a corresponding solution $\{y \to x,k\}$.
<br><br>
It follows that if $k \neq y \pm{1}$, then either $k \in K$ or $y \in K$.