In order the positive integer $n$ to have exactly 6 positive divisors, it must be of the form $n^2-9 = q^5$ or $n^2-9 = p^2q$, where $p$, $q$ are primes mutually different.

In the first case we have $n^2-9 = q^5 \Rightarrow (n-3)(n+3) = q^5$ (1)
It gives that
$$n-3 = q^s \text{ and } n+3 = q^t, \text{ with } t > s \text{ and } t+s = 5.$$
By subtracting we get $q^t - q^s = 6 \Leftrightarrow q^s(q^{t-s} - 1) = 2 \cdot 3$ and hence $q \in \{2,3\}$.
For $q=2$ (1) gives $n^2 = 41$, (it has no solutions in $\mathbb{Z}$), while for $q=3$, (1) gives $n^2 = 252$, (impossible in $\mathbb{Z}$).

In the second case $n^2-9 = p^2q$, it follows that $(n-3)(n+3) = p^2q$, and so
$$\begin{align*} (n+3 = pq \text{ and } n-3 = p) \text{ or } (n+3 = q \text{ and } n-3 = p^2) \\ \text{or } (n+3 = p^2 \text{ and } n-3 = q). \end{align*}$$
From the first case we get $p(q-1) = 6$, and hence ($p=3$ and $q=3$), which must be rejected because $p \neq q$.
From the last two cases we conclude that $\text{gcd}(n-3, n+3)=1$.