I have a question concering the family of Pellian equations $$x^2 - (k^2+1)y^2 = k^2. \qquad (*)$$

For an integer $k\geq 2$, the equation (*) has at least three classes of solutions in integers, corresponding to the fundamental solutions $(x_0,y_0) = (k, 0)$, $(k^2-k+1, k-1)$, $(k^2-k+1, -(k-1))$. Each fundamental solution induces a sequence (class) of solutions by $$ x+\sqrt{k^2+1}y = (x_0+\sqrt{k^2+1}y_0)(k+\sqrt{k^2+1})^{2m}. $$

For certain values of $k$, there are two additional classes of solutions. E.g. for $k=2t^2$, we have also fundamental solutions $(x_0,y_0) = (2t^3+t, \pm t)$, and additional solutions occur also for $k=4s^3-4s^2+3s-1$, for $k=F_{2n}$, and other polynomial or exponential subfamilies. However, I am not able to find any example such that (*) has more than five classes of solutions.

So, I am wondering does it make sense to state the conjecture that for $k\geq 2$, equation (*) always have exactly three or five classes of solutions (e.g. 3 or 5 fundamental solutions). Is there an obvious reason why should (or should not) the number of fundamental solutions of (*) be bounded by an absolute constant (independent on $k$)?

It is easy to see that each fundamental solution $(x_0,y_0)$ has to satisfy $|y_0| < k$, so the conjecture actually says that there is at most one solution of equation (*) with $0 < y < k-1$.

This question is related to the conjecture which says that there does not exist a set of four positive integer with the property that the product of any two of them is 1 greater than a square (see e.g. Section 3.1 of Diophantine m-tuples page and references given there).