Let $\alpha \in \mathbb{Q}_p^{\times}$. Then $\alpha \in \mathbb{Z}_p^{\times}$ iff there are infinitely many $n \in \mathbb{Z}_{>0}$ such that $X^n = \alpha$ has a solution.
Approach:
$\Rightarrow$ If $\alpha \in \mathbb{Z}_p^{\times}$, it is sufficient to show that there are infinitely many solutions in $\mathbb{Z} / p \mathbb{Z}$ by Hensel's lemma. Clearly for $n=k(p-1)+1$ (for all $k$), there is a solution, namely $\alpha$, since $\alpha^{k(p-1) +1} \equiv \alpha \mod p$. So there are infinitely many $n$'s. (Is my argument correct?)
$\Leftarrow$ Suppose $\alpha \not \in \mathbb{Z}_p^{\times}$. Then $ord_p(\alpha) \neq 0$. Since $X^n = \alpha$, we have $n \ ord_p(X) = ord_p(\alpha)$, so there can only be finetely many $n$'s such that there is a solution. (Correct?)
Can someone check my solution (or give an other proof)? I'am not sure of my proof, it looks too short. Thanks!.
When $p=2$ your argument fails: then $n=k+1$ so you are claiming that each element of $\Bbb Z_2^\times$ is a $n$-th power for all $n$. This is false: not all elements of $\Bbb Z_2^\times$ are squares (but they are all cubes).
It does suffice to exhibit just one $n>1$ for which each element in $\Bbb Z_p^\times$ is an $n$-th power. For $p$ odd, any $n$ coprime to $p(p-1)$ works.