On the Diophantine Equation (4^n)^x − P^y = Z^2

In this paper, the Diophantine equation (4n ) x − p y = z 2 , where p is an odd prime, n ∈ Z + and x, y, z are non-negative integers, has been investigated to show that the solutions are given by {(x, y, z, p)} = {(k, 1, 2 nk − 1, 2 nk+1 − 1)} ∪ {(0, 0, 0, p)}.