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.

$(1,1)$ is available by inspection.  Otherwise, if you write $2x^2=y^{15}+1=(y+1)(y^2-y+1)(y^4-y^3+y^2-y+1)(y^8+y^7-y^5-y^4-y^3+y+1)$ (Thanks, Alpha) you can look at where the factors of $2x^2$ can come from.  As the last three terms are odd, $y+1$ must have an odd numbers of factors of $2$. $y^4-y^3+y^2-y+1=(y^3-2y^2+3y-4)(y+1)+5$, so these two can share a 5, but no other prime.  $y^2-y+1=(y-2)(y+1)+3$ so these can share a factor $3$ but no other prime.  No other pairs can have common factors.  Aside from $2, 3, 5$, all primes must divide each factor an even number of times and can only divide one of them.  This gives you a lot of information on how the factors of $2x^2$ can be distributed.  I think it is pretty unlikely there are any more solutions
