How to find all the integral solutions of $x^2-by^2=z^k$ where $k$ is an odd integer >2 and $b>0$?

Question

Consider the bivariate quadratic polynomial of the form: $$ x^2-by^2=z^k$$ where $k$ is an odd integer>2 and $b>0$. It's well-known that Euler's method: $$x^2-by^2=(p^2-bq^2)^k $$ provides a class of integer solutions. I am interested in finding a method that provides all the integer solutions not given by Euler's . I was reviewing the collection of algebraic identities by @Tito Piezas. He mentioned the following alternate method: $$x^2-by^2=(p^2-bq^2)^k(r^2-bs^2)$$ with $r^2-bs^2=1$. My question still remains, do these methods provide all the possible solutions?

Answer 1

Dickson’s complete solution for the equation $x^2-my^2=zw$ should give all solutions.