Diophantine Equation in $\mathbb{Z}$

Question

I would like to know how to solve $2x^2 - y^{14} = 1$ in integers.

I've transformed it into $(y^7 - 1)^2 + (y^7 + 1)^2 = (2x)^2$ and I have stopped here.

Answer 1

Negate each side of the equation: \begin{align} y^{14} - 2x^2 = -1. \end{align} This is the negative Pell equation in $y^7$ and $x$, which is easy to find solutions for. Then simply look at the solutions for which $y$ is integral.

see Pell's equation for more information.