prove that the system
has no solutions in integers x, y and z.
The first equation is quadratic, so
So can be substituted in the second equation . After simplifying, we arrive at...
(Someone may want to check my algebra for small errors.)
Since the LHS is a perfect square, it would be sufficient to show that no integer values of y allow the RHS to be a perfect square. One approach may be to attempt to factor the RHS as a perfect square minus a small constant, and then show that there are only a finite number of squares who differ by that constant -- the smaller the better. This may be implausible by hand.
Hope this brute force progress helps!