You can factorize this polynomial like this: $$(n+1)(n^2+1)(n^4-n+1)$$ so this one has almost always at least 3 divisors.
Actualy, for $n>1$ we have $n+1>2$, $n^2+1>2$ and $n^4-n+1>2$ so it could be only $n=1$ which works.
How I got this this factorization:
\begin{eqnarray} (n^7+n^5)+(n^6+1) &=& n^5(n^2+1) + (n^2+1)(n^4-n^2+1)\\
&=& (n^2+1)(n^5+n^4-n^2+1)\\
&=& (n^2+1)\Big(n^4(n+1)-(n-1)(n+1)\Big)\\
&=&(n+1)(n^2+1)(n^4-n+1)
\end{eqnarray}
You do need Wilson's Theorem, but observe:
\begin{align}n &\equiv -(n-3) \pmod {2n-3}\\
n-1 &\equiv -(n-2) \pmod {2n-3}\\
&\ \vdots\\
1 &\equiv -(2n-4)\pmod {2n-3}\end{align}
This gives:
\begin{align}(n!)^2&\equiv n!(-1)^n(2n-4)(2n-5)\dots(n-2)(n-3) \pmod {2n-3}\\&\equiv(-1)^n(2n-4)!(n)(n-1)(n-2)(n-3) \pmod {2n-3}\\\text{(Wilson)} &\equiv(-1)^{n+1}(n)(n-1)(n-2)(n-3)\pmod {2n-3}\\
&\equiv(-1)^{n+1}n^2(n-1)^2 \pmod {2n-3}\end{align}
Now we seperate into cases where $n$ is odd or even. For $n$ odd satisfying the division condition,
$$15(n!)^2+1\equiv 15n^2(n-1)^2+1 \equiv 0 \pmod {2n-3}$$
$z = \dfrac {15n^2(n-1)^2+1}{2n-3}$ is an integer iff $16z$ is. Simplifying we have
$$16z=120 n^3 - 60 n^2 + 30 n + 45 + \frac {151} {2 n - 3}$$
So the above is an integer only if $2n-3$ divides $151$, which is a prime, giving $2n-3 = 151$, $n = 77$.
The case for $n$ even should be similar. Technically, you also need to consider the cases $2n-3 = \pm 1$ separately, since $\pm1$ are neither prime nor composite.
Best Answer
I think one has to include $0$ as a natural number for this question, as any $N$ has the natural divisor $1,$ which decreased by $1$ gives $0,$ which is the square of $0$ (but not the square of a positive nnatural (some say only positive integers are natural numbers). Let's put that triviality aside-- Now you're looking for non-prime examples, so consider $N=10,$ whose (non-1) divisors are $2,5,10$ and these decreased by $1$ are all squares, $1,4,9.$
Note that this example is a product of primes to the first power, as the OP comments in the post that he's shown (somehow). BTW that proof would be appreciated as an add-on to the question, at least by me.
added: I think it's not hard to show that one cannot have other $N$ than $10$ which are even and divisible by some other odd prime $p$, by considering only the facts that one would need both $p=x^2+1$ and $2p=y^2+1,$ among whatever other equations if there are more prime divisors. One gets to $y=x+1$ and keeps going. So what's left is the case of odd $N.$ [That's if the last paragraph works out for a proof of that case.]