Can we find solutions of Diophantine equations of the form :
$$(2^n)^x + p^y = z^2 $$
where $k, x, y, z$ and $n$ are positive integers.
Some easily-spotted solutions:
Another simple solution:
x = 1, p = 2, n even, y = n+3
Then, putting m = n/2:
$(2^n)^x + p^y = 2^n + 2^3.2^n = (1 + 2^3).2^n = 9.2^n = 9.4^m = (3.2^m)^2$