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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 $$2k = A^2 B^2 - (k^2+1)B^2. k^2+1)A^2.$$ The solution $B A = 1$ and $A B = k+1$ 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?

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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 $$2k = A^2 - (k^2+1)B^2.$$ The solution $B = 1$ and $A = k+1$ 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?