Case 1: $n^2-9=p^5$ for a prime $p$. We have $(n-3)(n+3)=p^5$. Then either $n-3$ is 1, $p$, or $p^2$. In the first case, $n=4$, but $n^2-9=7$ does not have 6 divisors. In the second case, we have $p^4-p=6$, but this has no integer solutions for $p$. In the third case, $p^3-p^2=6$, which also has no integer solutions for $p$. Case 2: $n^2-9=pq^2$ for distinct primes $p$ and $q$. The possibilities for $n-3$ are 1, $p$, $q$, $q^2$. We only need to verify that the third case is not possible. So suppose that $n-3=q$, so that $n+3=pq$. Then $p=\frac{n+3}{n-3}=1+\frac{6}{n-3}$ is an integer exceeding 1. Since $n-3$ divides 6, the possibilities for $n$ are 4,5,6, and 9. However, testing each of these yields $n^2-9$ to be 7, 16, 27, and 72 respectively, none of which has exactly 6 divisors. Thus, $\gcd(n-3,n+3)=1$.

If $n=2k+1$ then $n^2-9=4k^2+4k-8$. We see that for $k>2$, atleast $7$ factors of $n^2-9$ are distinct .Now checking manually for $n=1,3,5$, we do not get $6$ distinct divisors in each of the $3$ cases. So, $n$ must be even. Let $\gcd(n+3,n-3)=k\implies k|6\implies k=1,2,3,6$, but $k\neq 2,6$ as $n$ is even. So, $k=1,3$. If $k=3\implies 3|n$ also $2|n\implies 6|n$. Let $n=6m\forall m\in\mathbb Z^+$. Notice that for $m>1, n^2-9$ has at least distinct $7$ divisors and for $m=1, n^2-9=27$ which doesn't have $6$ divisors. Hence, $k=1$. $\blacksquare$.

amar_04 wrote: If $n$ is odd say $n=2k+1$, then $n^2-9$ has at least $7$ divisors. This statement is wrong. A counter-example is $n=5$, fo which $n^2-9=16$ has only $5$ divisors $1,2,4,8,16$.

amar_04 wrote: Let $n=3m$ for some $m\in\mathbb Z^+$, hence $n^2-9$ has more than $6$ divisors. And that statement is also wrong! A counter-example is $n=6$, for which $n^2-9=27$ has only $4$ divisors $1,3,9,27$.

prague123 wrote: amar_04 wrote: If $n$ is odd say $n=2k+1$, then $n^2-9$ has at least $7$ divisors. This statement is wrong. A counter-example is $n=5$, fo which $n^2-9=16$ has only $5$ divisors $1,2,4,8,16$. Sorry, my internet connection went away, when I was actually checking my solution, I've edited it, hopefully it's correct now.

Suppose the $ gcd (n-3,n+3)>1 $ and $ n-3=dx_1;n+3=dx_2; (x_1,x_2)=1 $.Therefore $ n^2-9=d^2.x_1.x_2 $ and $ d (x_1-x_2)=6 $.Because $ n^2-9 $ has 6 devisors, obviously d is prime or power of prime(otherwise d has at least 4 devisors, and $ x_1.x_2 $ has at least 2 devisors and because $ (d,x_1,x_2)=1 $ the condition in the statement is impossible). Therefore d=2 or 3. In the both cases $ x_1.x_2 $ must have 2 devisors and because $ x_1-x_2=3 $ or $ 2 $, $ x_2=1 $ and $ x_1=4 $ or $ 3 $ which obviously contradicts.