On the Diophantine equation $(2^x-1)(p^y-1)=2z^2$

summary:Let $p$ be an odd prime. By using the elementary methods we prove that: (1) if $2\nmid x$, $p\equiv \pm 3\pmod 8,$ the Diophantine equation $(2^{x}-1)(p^{y}-1)=2z^{2}$ has no positive integer solution except when $p=3$ or $p$ is of the form $p=2a_{0}^{2}+1$, where $a_{0}>1$ is an odd positive integer. (2) if $2\nmid x$, $2\mid y$, $y\neq 2,4,$ then the Diophantine equation $(2^{x}-1)(p^{y}-1)=2z^{2}$ has no positive integer solution.