solving a strange Diophantine equation ${\sqrt{n}}^\sqrt{n} -11 =m!^2$ 
Does anyone know how to solve Diophantine equation: $${\sqrt{n}}^\sqrt{n}-11 =m!^2.$$

I tried to substitute $\sqrt{n}=k$ then equation becomes $$k^k-11=m!^2\\\implies k^k=m!^2+11=(m!-1)(m!+1)+12$$ which means suppose $m\geq2$ then $\gcd{(m!-1,m!+1)}=1$. Does this give any hint? I could think upto here only. Please help.

Let
$$k^k-11 = (m!)^2.$$
If $k>=11$ then $$121\not|\, LHS,\quad 121\,|\, RHS,\quad LHS\not= RHS.$$
In the other hand, $k$ is odd and $k^k > 11$, so
$$k\in\{3,5,7,9\}.$$
Note than:
$$\sqrt{3^3-11} = 4 \not= m!,$$
$$\sqrt{5^5-11} = \sqrt{3114}\in(55,56),\quad \sqrt{5^5-11}\not\in\mathbb N,$$
$$\sqrt{7^7-11} = \sqrt{823532}\in(907,908),\quad \sqrt{7^7-11}\not\in\mathbb N,$$
$$\sqrt{9^9-11} = \sqrt{387420478}= \sqrt{11683^2-1}\in(11682,11683),\quad \sqrt{9^9-11}\not\in\mathbb N.$$
So the issue diophantine equation has not solutions.
