$S_4$ does not have a normal subgroup of order $8$

You approach is OK, but here is a simpler one. Since $S_4/H \cong C_3$ is abelian, it follows that $[S_4,S_4]=A_4 \subseteq H$. Can you see this leads to a contradiction?

Yet another approach: Normal subgroups are comprised of entire conjugacy classes, one of which must be the identity's conjugacy class, but the class equation for $S_4$ is $$24=1+3+6+6+8,$$ and there is no way to obtain $8$ as a sum of terms from the right hand side including $1$.

Another approach : Two elements in $S_n$ are conjugate to each other iff they have the same cycle decomposition. Hence, all the transpositions must be in $H$ if one of them is (because $H$ is normal).

This leaves one non-identity element, which must have order 2 (since all the transpositions are their own inverses). That element must be a product of transpositions. Such a thing would be conjugate to something else, which is not in $H$.

If you assume that $H$ does not contain any transpositions, it still contains an element of order 2 ...