On the Diophantine equation $x^{p}+2^{2m}=py^{2}$

Let $p$ be an odd prime. In this paper, using some theorems of Adachi and the author, we prove that if $p \equiv 1(\text{mod }4)$ and $p\nmid B_{(p-1)/2}$, then the equation $x^{p}+1=py^{2},\,\,y\ne 0$, and the equation $x^{p}+2^{2m}=py^{2},\,\,m\in \mathbb{N} , \text{ gcd}(x, y )=1,\,\, p\mid y$, have no integral solutions respectively. Here $B_{(p-1)/2}$ is $(p-1)/2$th Bernoulli number.