We have $p^3\mid k!$ thus $k\ge 3p$. So \begin{align*} (3p+m)! &=p^3(p^3-1)(p^2-1)(p-1) \\ \prod_{i=1}^{p-1} (2p+i) &<(p^3-1)(p^2-1)(p-1) \\ \end{align*}If $p\ge 7$ we have $\prod_{i=1}^6 (2p+i)<(p^3-1)(p^2-1)(p-1)<p^6$ but obviously $p^6<\prod_{i=1}^6 (2p+i)$ so $p<7$. If $p=5$ we have $31\mid k!$ so $k\ge 31$ which is obviously a size contradiction. If $p=3$ we have $13\mid k!$ so $k\ge 13$ but so $k!$ is larger than $RHS$. If $p=2$ we have $7\mid k!$ thus $k\ge 7$. Moreover $7!>RHS$. Thus there are no solutions.

We claim that there exists no such solutions. Now, assume there exists a pair of solutions $(k,p)$ for which the required condition is satisfied. Then, \[\nu_p(k!)=\nu_p((p^3-1)(p^3-p)(p^3-p^2))=3\]Thus, $k\geq 3p$. Now, we consider a prime $q\neq 2,3$ such that $q\mid p-1$. Then, \begin{align*} \nu_q(p^3-1) &= \nu_q(p-1)+\nu_q(p^2+p+1)\\ &= \nu_q(p-1)\\ \nu_q(p^3-p) &=\nu_q(p)+\nu_q(p-1)+\nu_q(p+1)\\ &= \nu_q(p-1)\\ \nu_q(p^3-p^2) &= 2\nu_q(p)+\nu_q(p-1)\\ &= \nu_q(p-1) \end{align*}Thus, we are able to conclude that, \[\nu_q((p^3-1)(p^3-p)(p^3-p^2)=3\nu_q(p-1))\]But, for all prime $q\neq 2$, we have that $\nu_q\left(\frac{p-1}{2}\right)=\nu_q(p-1)$. Thus, \[\nu_q(k!) \geq \nu_q\left(\frac{p-1}{2}\right) + \nu_q(p-1) + \nu_q\left(\frac{3p-3}{2}\right) + \nu_q(2p-2) + \nu_q\left(\frac{5p-5}{2}\right) + \nu_q(3p-3) \geq 6 \nu_q(p-1)\]since, $k\geq 3p$ as mentioned before. But this means, \[\nu_q(k!) \geq 6\nu_q(p-1) > 3\nu_q(p-1) = \nu_q((p^3-1)(p^3-p)(p^3-p^2))\]which is a clear contradiction. Further, assume $\nu_3(p-1) \geq 2$. \begin{align*} \nu_3(p^3-1) &= \nu_3(p-1)+\nu_3(p^2+p+1)\\ &= \nu_3(p-1)+1\\ \nu_3(p^3-p) &=\nu_3(p)+\nu_3(p-1)+\nu_3(p+1)\\ &= \nu_3(p-1)\\ \nu_3(p^3-p^2) &= 2\nu_3(p)+\nu_3(p-1)\\ &= \nu_3(p-1) \end{align*}But, we have that \[\nu_3(k!)\geq 6\nu_3(p-1) > 3\nu_3(p-1)+1 = \nu_3((p^3-1)(p^3-p)(p^3-p^2))\]Thus, $\nu_3(p-1)\leq 1$. Similarly assume $\nu_2(p-1)\geq 2$. Then, \begin{align*} \nu_2(p^3-1) &= \nu_2(p-1)+\nu_2(p^2+p+1)\\ &= \nu_2(p-1)\\ \nu_2(p^3-p) &=\nu_2(p)+\nu_2(p-1)+\nu_2(p+1)\\ &= \nu_2(p-1)+1\\ \nu_2(p^3-p^2) &= 2\nu_2(p)+\nu_2(p-1)\\ &= \nu_2(p-1) \end{align*}Here, note that, \[\nu_2\left(\frac{p-1}{2}\right)=\nu_2(p-1)-1\]But this means, \[\nu_2(k!) \geq 6\nu_2(p-1)-3 > 3\nu_2(p-1)+1 = \nu_2((p^3-1)(p^3-p)(p^3-p^2))\]Thus, $\nu_2(p-1)\leq 1$. This means, $p-1\in \{1,2,3,6\}$. Thus, $p\in \{2,3,4,7\}$. Checking each of these possibilities separately, we see that there exist no solutions as expected.