We must find all positive integers $n$ such that $\left(n ^2+11n-4\right)n!+33\times13^n+4$ is a perfect square. Let $m=\left(n ^2+11n-4\right)n!+33\times13^n+4$ be this integer. We will cut this problem in two cases and use quadratic residue. Assume $m$ is a perfect square. Case 1: Let $n\geqslant7$. We know that the quadratic residue $\pmod{7}$ are element of $\left\{0,1,2,4\right\}$. Since $n!\equiv0\pmod{7}$ we can write \begin{align*} m&=\left(n ^2+11n-4\right)n!+33\times13^n+4\\ &\equiv-2\times\left(-1\right)^n+4\pmod{7} \end{align*}which gives $m\equiv2\pmod{7}$ iff $n$ is even or $m\equiv6\pmod{7}$ iff $n$ is odd. Since $6$ is not a quadratic residue $\pmod{7}$ we must have $n$ even. So, let's take $n=2k$ (with $k$, a positive integer). Now, since $n!\equiv0\pmod{8}$ (as $n\geqslant7$) and the quadratic residue $\pmod{8}$ are element of $\left\{0,1,4\right\}$ we have \begin{align*} m&=\left(n ^2+11n-4\right)n!+33\times13^n+4\\ &\equiv5^n+4\pmod{8}\\ &\equiv(5^2)^k+4\pmod{8}\\ &\equiv5\pmod{8} \end{align*}but this means that $m$ is not a quadratic residue $\pmod{8}$. Thus, if $n\geqslant7$ then $n$ can't be both a perfect square $\pmod{7}$ and $\pmod{8}$. This is a contradiction so $n<7$. Case 2: Let $n<7$. Then casework shows that only $n=1$ and $n=2$ leads to a perfect square. In conclusion, the only positive integers that are solution are $\boxed{\left(n,m\right)=\left\{\left(1,21^2\right),\left(2,75^2\right)\right\}}$.

We must find all positive integers $n$ such that $\left(n ^2+11n-4\right)n!+33\times13^n+4$ is a perfect square. Let $m=\left(n ^2+11n-4\right)n!+33\times13^n+4$ be this integer. We will cut this problem in two cases and use quadratic residue. Assume $m$ is a perfect square. Case 1: Let $n\geqslant7$. We know that the quadratic residue $\pmod{7}$ are element of $\left\{0,1,2,4\right\}$. Since $n!\equiv0\pmod{7}$ we can write \begin{align*} m&=\left(n ^2+11n-4\right)n!+33\times13^n+4\\ &\equiv-2\times\left(-1\right)^n+4\pmod{7} \end{align*}which gives $m\equiv2\pmod{7}$ iff $n$ is even or $m\equiv6\pmod{7}$ iff $n$ is odd. Since $6$ is not a quadratic residue $\pmod{7}$ we must have $n$ even. So, let's take $n=2m$ (with $m$, a positive integer). Now, since $n!\equiv0\pmod{8}$ (as $n\geqslant7$) and the quadratic residue $\pmod{8}$ are element of $\left\{0,1,4\right\}$ we have \begin{align*} m&=\left(n ^2+11n-4\right)n!+33\times13^n+4\\ &\equiv5^n+4\pmod{8}\\ &\equiv(5^2)^m+4\pmod{8}\\ &\equiv5\pmod{8} \end{align*}but this means that $m$ is not a quadratic residue $\pmod{8}$. Thus, if $n\geqslant7$ then $n$ can't be both a perfect square $\pmod{7}$ and $\pmod{8}$. This is a contradiction so $n<7$. Case 2: Let $n<7$. Then casework shows that only $n=1$ and $n=2$ leads to a perfect square. In conclusion, the only positive integers that are solution are $\boxed{\left(n,m\right)=\left\{\left(1,21^2\right),\left(2,75^2\right)\right\}}$.