TITLE: Diophantine Equation in $\mathbb{Z}$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.