After substituting x=z^2 and y=az^2 you can divide by z^2 and get

z^2(a^2 + 1) + 1 = 2kaz^2 which is possible only if z=1.

Then x=1 and y=a. Plugging in these values gives a quadratic in 'a' with discriminant 4(k^2 - 2).

But k^2 - 2 = 2 or 3 mod4, so will never be a square.

Thus, there are no solutions.