The equation can be re-written lilke this : $(m+1)(m-1) = 5.2^n$
Notice that either $(m+1)$ or $(m-1)$ can not be $5$.Because if $(m+1)$ or $(m-1)$ is $5$ then the other will also be an odd number.Thus,the L.H.S is odd.But the R.H.S. is even and so it's impossible.
So, we can say that $m+1 = 2^k$ or $m-1 = 2^k$ where $k$ is a natural number.
Case 1:
If,$m-1 = 2^k$
then , $m+1 = 2^k + 2$
Moreover , $m+1 = 5.2^q$ where $q$ is another natural number.
So,$2^k + 2 = 5.2^q \rightarrow 2(2^(k-1) + 1) = 5.2^q$
$2^(k-1) + 1$ is odd.But $5.2^q$ has only one odd factor which is $5$
Thus,$2^(k-1) + 1 = 5$ or, $k = 3,m = 9 , n = 4$.
Case 2:
If,$m+1 = 2^k$
then , $m-1 = 2^k - 2$
Moreover , $m-1 = 5.2^p$ where $p$ is another natural number.
So,$2^k - 2 = 5.2^p \rightarrow 2(2^(k-1) - 1) = 5.2^p$
Similar to the first case,$2^(k-1) - 1 = 5$
But there is no such $k$.And so,there is no solution in this case.
Therefore,we get $(m , n) = (9 , 4)$