I'm looking for solutions to an equation of form $m =\sqrt{x^2 - 8y^2}$. I know that $m$ is a positive integer and so the inside of the square root has to be complete square. So I'm stuck with this diophantine equation $x^2 - 8y^2 = m^2$. In the case where $m = \pm1$ that's Pell Equation which I know how to find solutions to. I also know that I can find infinitely many solutions because solutions to $x^2 - 8y^2 = 1$ are also solutions to $(mx)^2 - 8(my)^2 = m^2$ but that doesn't capture all solutions.
This site seems to derive the solutions by solving $s^2 - 8 = 0$ mod $m^2$ but I don't know why and I don't know what to do with it.
Most papers I find on the topic say they have a solution if $\sqrt{8} > m^2$ but that's not sufficient here.
For example: $m = 7$ -> $x^2 - 8y^2 = 49$ and I want to be able to arrive at $x = 11; y = 3$ and other solutions where $x$ and $y$ aren't $0$ mod $7$.
Any help is appreciated. Thank you.
