In this paper, we gave solutions of the Diophantine equations
16^{x}+p^{y}=z^{2}, 64^{x}+p^{y}=z^{2} where p is an odd prime, n is a positive
integer and x,y,z are non-negative integers. Finally we gave a generalization
of the Diophantine equation (4^{n})^{x}+p^{y}=z^{2}.
In this article, we consider the equation x^2+19^{m}=y^n, n>2, m>0. We find
the solutions of the title equation for not only 2 \mid m but also
2\notdividesm.