TL;DR: Clifford hierarchy is not closed under raising to integer powers. One suitable counterexample is $g:=TH$ which resides in the third level $\mathcal{C}^{(3)}$ of the Clifford hierarchy, but whose square $g^2=THTH$ is altogether outside the hierarchy $\mathcal{C}=\cup_{i=1}^\infty\mathcal{C}^{(i)}$.
Right-multiplication by Cliffords fixes high levels of the hierarchy
The first level $\mathcal{C}^{(1)}$ of the $n$-qubit Clifford hierarchy is the $n$-qubit Pauli group and the $k$th level is defined recursively as
$$
\mathcal{C}^{(k)}:=\{U\in U(2^n)\,|\,\forall P\in\mathcal{C}^{(1)}\quad UPU^\dagger\in\mathcal{C}^{(k-1)}\}.\tag1
$$
In particular, conjugation by any Clifford $C\in\mathcal{C}^{(2)}$ sends any Pauli operator $P\in\mathcal{C}^{(1)}$ to a Pauli operator $Q\in\mathcal{C}^{(1)}$
$$
CPC^\dagger=Q\in\mathcal{C}^{(1)}.\tag2
$$
Similarly, every $U\in\mathcal{C}^{(k)}$ for $k\geqslant 2$ sends every Pauli operator $Q\in\mathcal{C}^{(1)}$ to a unitary in $\mathcal{C}^{(k-1)}$
$$
UQU^\dagger=V\in\mathcal{C}^{(k-1)}.\tag3
$$
Combining $(2)$ and $(3)$, we see that for every Clifford $C\in\mathcal{C}^{(2)}$ and every $U\in\mathcal{C}^{(k)}$
$$
UCP(UC)^\dagger=UCPC^\dagger U^\dagger=UQU^\dagger=V\in\mathcal{C}^{(k-1)}.\tag4
$$
Thus, we obtain the equivalence
$$U\in\mathcal{C}^{(k)}\iff UC\in\mathcal{C}^{(k)}\tag5$$
for every Clifford $C\in\mathcal{C}^{(2)}$ and any integer $k\geqslant 2$.
Third level is not closed under squares
Finally, $T\in\mathcal{C}^{(3)}$, so by $(5)$ we have $g:=TH\in\mathcal{C}^{(3)}$. But, we know that $THT$ is not in the Clifford hierarchy $\mathcal{C}$, so using $(5)$ again we obtain$^1$ that $g^2=THTH\notin\mathcal{C}$. Thus, $g$ is an operator in the third level of the Clifford hierarchy whose square is not in any level of the hierarchy.
$^1$ The equivalence in $(5)$ only implies that $g^2$ is not in the second or higher level of the Clifford hierarchy, but $THTH$ is a $2\times 2$ matrix without zero elements, so it is not in $\mathcal{C}^{(1)}$, either.
