Is not it impossible to have $p^{2l-1}m(mn+1)^2$ a perfect square when $m$ is a perfect square? Since, the exponent of $p$ in the expression must be odd?

Sorry, there was a typo. I fixed it.

The number is $m\left (p^{2\ell-1}(mn+1)^2 + m\right )$. Say $q\mid m$, $q\neq p$, $q$ prime; then $\gcd\left (p^{2\ell-1}(mn+1)^2 + m,q\right )=1$, so $q$ must appear at an even exponent in $m$. Say $p\mid m$. If $p$ appears at an even exponent in $m$, we are done, so assume $m = p^{2k-1}m'^2$ for some positive integer $k$, where $p\nmid m'$. Then $p\left (p^{2\ell-1}(mn+1)^2 + m\right ) = \left (p^\ell(mn+1)\right )^2 + pm$ must be a perfect square. But $\left (p^\ell(mn+1) + 1\right )^2 = \left (p^\ell(mn+1)\right )^2 + 2p^\ell(mn+1) + 1 > \left (p^\ell(mn+1)\right )^2 + pm > \left (p^\ell(mn+1)\right )^2$, so that is impossible.