All natural solutions of $2x^2-1=y^{15}$ How can I find all positive integers $x$ and $y$ such that $2x^2-1=y^{15}$?
PS. See here.

I was working on Byron's suggestion before he made it but it took me a while because I'm not a real number theorist.  And, I used Sage.  I consider $2y^2 = x^3 + 1$ and I want to put this in a form where the coefficient of $y^2$ and $x^3$ are 1 so I multiply both sides by 1/2 first and then use the transform $(x,y) \mapsto (X/2,Y/4)$ to get $Y^2 = X^3 + 8$.  Now, Sage can find all integral points and they are $(-2, 0)$, $(1, \pm3)$, $(2, \pm4)$, and $(46, \pm312)$.  This is good enough because note that any integer solution to the original $2y^2 = x^3 + 1$ will be mapped to an integer solution here.  So, now you just need to consider these 7 solutions.  Our map is clearly invertible by $(X, Y) \mapsto (2x, 4y)$, so we have a one-to-one correspondence between points and everything makes sense.  All we need to do is map those 7 points backward to see what they were on $2y^2 = x^3 + 1$.  They were $(-1, 0)$, $(1/2, \pm 3/4)$, $(1, \pm1)$, and $(23, \pm78)$.  Thus, we have all integer solutions to $2y^2 = x^3 + 1$, 3 are obvious and the other two, $(23, \pm78)$ can not correspond to solutions of $2y^2 = x^{15} + 1$ as Byron explained.
By the way, the elliptic curve has rank 1 which means it has an infinite number of solutions over the rationals.  But, whether or not any of these correspond to nontrivial solutions of the original $2y^2 = x^{15} + 1$, I have no idea.
