Is the occurence of two perfect numbers a coincidence? The expression $$n^n+n!+1$$ is prime for the following non-negative integers $n\le 7\ 600$ : $$[0, 1, 2, 4, 28, 496]$$ if we assume $0^0=1$
The numbers $28$ and $496$ are perfect numbers.

  
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*Is this just a coincidence or is there an explanation ?
  
*Do further primes of the given form exist ?
  

$1$ is the only odd possible $n$
Proof : Suppose $n\ge 3$ is odd and $p$ is a prime factor of $n+1$. Then clearly $p\le n$ (because $n+1\ge 4$ is even, hence no prime) , hence $p|n!$ and we have $n^n\equiv (-1)^n\equiv -1\mod p$ , ruling out odd numbers $n\ge 3$
What about the even numbers $n$ ? Are there additional conditions explanating the perfect numbers and accelerating the search for further primes ?
 A: Isn't an answer, just to add a remark. 
When we assume that $n\geq 1$ (I exclude of this discussion the case $n=0$) is a prime number, from Fermat's little theorem one has that $$2^{n^n+n!}\equiv 1\text{ mod } n^n+n!+1,\tag{1}$$
thus there exists a positive integer $k$ (depending on our fixed integer $n$, that is $k=k(n)$) such that $$2^{n^n+n!}=k\cdot(n^n+n!+1)+1.\tag{2}$$
Previous paragraph is the preparation to invoke the so-called abc conjeture, that as you know is an important conjecture in analytic number theory.
Invoking the abc conjecture (I am saying the ABC Conjecture II from the Wikipedia's page) $\forall\epsilon>0$ there exists a positive constant $C(\epsilon)$ (having some well-known and important properties) such that $$2^{n^n+n!}<C(\epsilon)\operatorname{rad}\left(2^{n^n+n!}\cdot k\cdot(n^n+n!+1)\right)^{1+\epsilon}.\tag{3}$$
Since you know that the arithmetic function $\operatorname{rad}(m)$ satisfies for integers $e,f\geq 1$ the obvious inequality $$\operatorname{rad}(ef)\leq \operatorname{rad}(e)\operatorname{rad}(f)$$
(and also always $\operatorname{rad}(m)\leq m$) one deduces from $(3)$ the inequality $$2^{n^n+n!}<C(\epsilon)\cdot 2^{1+\epsilon}(n^n+n!+1)^{1+\epsilon}\cdot \operatorname{rad}(k)^{1+\epsilon}.\tag{4}$$
