Solutions to simultaneous Diophantine equations $2y^2-3x^2=-1$ and $z^2-2y^2= -1$

Question

I am looking for integer solutions for the following set of equations:

$2y^2-3x^2=-1$

$z^2-2y^2= -1$

I know that there are the solutions (1,1,1) and (-1,-1,-1) for this set of simultaneous equations. How would I go about showing that there are no other solutions to these simultaneous equations?

Answer 1

There are no integer solutions other than the trivial $\lvert x \rvert = \lvert y \rvert = \lvert z \lvert = 1.$

An elementary solution can be found at this earlier (duplicate) post.