On the Diophantine Equation $p^x+(2p+1{)}^y=z^2$ with Consecutive Exponents

We study the Diophantine equation $p^x+(2p+1{)}^y=z^2$ over positive integers $x$, $y$ and $z$ for every odd prime $p$. We prove that $(x,y,z)=(2,1,p+1)$ is the unique solution except possibly when $2p+1$ is composite. In that case, it reduces to a family depending on one parameter, and we show that the parameter must be odd and satisfies an explicit upper bound, reducing the problem to finitely many cases.