We shall find a pair such that $m+n=p$ is prime and $n$ is even. Applying Wilson's theorem we have

$$m!=(p-n)!=\frac{(p-1)!}{(p-n+1) \ldots(p-2)(p-1)} \equiv \frac{-1}{-(n-1) \ldots(-2)(-1)} \equiv \frac{1}{(n-1)!} \equiv \frac{n}{n!} \quad(\bmod p)$$

It follows from Fermat's Little Theorem that $(n!)^{p} \equiv n!(\bmod p)$, therefore

$$(m!)^{n}+(n!)^{m}+1 \equiv\left(\frac{n}{n!}\right)^{n}+(n!)^{p-n}+1 \equiv \frac{n^{n}+n!+(n!)^{n}}{(n!)^{n}} \quad(\bmod p)$$

thus it suffices to prove that the number $n^{n}+n!+(n!)^{n}$ has a prime divisor $p>n$ for infinitely many even $n$.

We prove that this condition is satisfied, for instance, by all the numbers of the form $n=2 q$, where $q>2$ is prime. Let $A=(2 q)^{2 q}+(2 q)!+((2 q)!)^{2 q}$. For a prime $p$ and integer $k$ we denote by $v_{p}(k)$ the largest integer $\ell$ such that $p^{\ell}$ divides $k$.

If $r(2 q)^{2 q}>2^{2 q-1} q^{2}$, therefore $A$ has a prime divisor $p>2 q$, q.e.d.