A number theory problem by Snehashis Mukherjee

$n^2-4=5 3^m$

$(n+2)(n-2)=5 3^m$

Case(1) : $n+2=5.3^r$ and $n-2=3^{m-r}$

Subtracting these two equations : $5.3^r-3^{m-r}=4$

L.H.S. part will give remainder zero when it is divided by $3$ but R.H.S. will give remainder 1 when divided by $3$ .So these combination is not possible

Case(2) : $n+2=3^r$ and $n-2=5.3^{m-r}$

These part will also not possible same reason as in Case(1).

Case(3): $n+2=3^m$ and $n-2=5$

Subtracting it will give $3^m-5=4 \\ 3^m=9 \\ m=2$

So, $n=7$

Case(4): $n+2=5$ and $n-2=3^m$

Subtracting we get $5-3^m=4 \\ 3^m=1 \\ m=0$

So $n=3$

Total possible values of $n= \boxed{10}$