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. <a href="http://web.math.pmf.unizg.hr/~duje/dn.html" title="Section 3.1 of Diophantine m-tuples page">Section 3.1 of Diophantine m-tuples page</a>
and references given there).