Here is a hint. Find all the cycle types of permutations of $23$ (i.e., all partition of $23$) such that the fourth power of a permutation of that type gives a cycle type $(3,3,3,3,3,2,2,2,2)$. Instead of looking at all partitions of $23$, you can rapidly narrow down the search for such partitions by looking at what happens to individual cycles when taking fourth powers. Once you have the list, use that the number of permutations with type a partition $\lambda=(\lambda_1,\ldots,\lambda_k)$ of $n$ is given by $ \frac{n!}{(\prod_{i=1}^k\lambda_i)(\prod_{l=1}^{\lambda_1}m_l(\lambda)!)} $ where $m_l(\lambda)$ counts the number of parts of $\lambda$ equal to $l$. For instance for $\lambda=(8,6,3,3,3)$ (which should appear on your list) you get $ \frac{23!}{(8\times6\times3\times3\times3) (0!\times0!\times3!\times0!\times0!\times1!\times0!\times1!)} =3\,324\,590\,630\,000\,640\,000 $ different permutations.
Oops, I forgot, this will give you the number of all permutations whose fourth power is some permutation of type $(3,3,3,3,3,2,2,2,2)$. To get the number of permutations giving a specific such permutation $\tau$, divide by the total number $2308743493056000$ of such permutations. The division must be exact, which gives a nice test for errors. Maybe it is easier (and avoids large numbers better) to figure out for every $\lambda$ separately how many possibilities you have for a permutation $\sigma$ of type $\lambda$ to satisfy $\sigma^4=\tau$.