So I'm working on the Diophantine equation $2x^2-1=y^{15}$ (1) with $x,y>1$
In particular I want to show that x must be a multiple of 5. I have found that it suffices to show that for $y=1 \pmod{10}$ (1) can't have a solution, but I'm stuck.
Also (I don't know if this helps) $y^{15}+1$ can be written as $(y^5+1)(y^2-y+1)(y^8+y^7-y^5-y^4-y^3+y+1)$
Here's an idea. Rewrite the equation as
$$16x^2=8y^{15}+8$$
and then let $v=4x$ and $u=2y^5$. This turns things into an elliptic curve equation
$$v^2=u^3+8$$
which, if I recall correctly, can have at most finitely many integer solutions. Not all integer solutions $(u,v)$ correspond to integer solutions $(x,y)$, but all integer solutions $(x,y)$ do correspond to integer solutions $(u,v)$.