a family of Pellian equations 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).
Edit (2.10.2022): In the recent paper by Maohua Le and Anitha Srinivasan, the conjecture is verified for $k=p^m q^n$, where $p,q$ are distinct odd primes and $m,n$ are positive integers (see https://web.math.pmf.unizg.hr/glasnik/forthcoming/pGM7053.pdf).
 A: This is an interesting question. Let me explain why I believe that the answer is no.
Rewrite the equation in the form 
$$ (x-k)(x+k) = (k^2+1)y^2. $$
Interpreting this as an equation in the polynomial ring ${\mathbb Z}[k]$ and using unique factorization you end up with equations of the form 
$$ x - k = (k^2+1)A^2, \quad x + k = B^2, $$
which implies
$$ 2k = B^2 - (k^2+1)A^2. $$
The solution $A = 1$ and $B = k+1$ gives rise to one of yours.
In order to be able to find a lot more fundamental solutions you will have to substitute $k = f(t)$ for a polynomial $f$ that makes $k^2+1$ reducible. the simplest choice is $k = 2t^2$ since 
$$ k^2+1 = 4t^4 + 1 = (2t^2+1)^2 - 4t^2 = (2t^2+2t+1)(2t^2-2t+1). $$
Going through the game above will reveal your additional solution in this case.
If you substitute $ k = t + 2t^3$ (I hope I remember my calculations well), the polynomial $k^2+1$ splits into three quadratic factors. Whether the resulting Pell type equations have more solutions is difficult to check. But my impression is that of you can find many substitutions for which $k^2+1$ splits into many factors, chances are that you get more than 2*2+1 = 5 fundamental solutions. Where's Noam when you need him? 
