Given $x\in\mathbb{Q}_p^*$, we can write $x=p^nu$ where $n\in\mathbb{Z}$ and $u\in\mathbb{Z}_p^*$. Then we can decide whether or not $x$ is a square by looking at $n$ and $u$. If $p\neq 2$ then $x$ is a sqaure if and only if the "first component" of $x$ is a square as a member of $\mathbb{F}_p^*$ and $n$ is even. If $p=2$ then $x$ is a square if and only if $n$ is even and $u\in1+8\mathbb{Z}_p$.
Is there a similar characterization of when $x\in\mathbb{Q}_p^*$ is an $m$-th power?
A necessary condition is $m\mid n$ and $u$ is an $m$-th power in $\mathbb F_p$. So the real question is:
Let $a\in \mathbb Z_p$. When is $1+pa$ an $m$-th power in $\mathbb Z_p$ ?
Edit The suggestion of Jyrki in the comments makes me think to the general solution. Let $d=v_p(m)$. Then:
Claim :
If $p>2$, then $1+pa$ is an $m$-th power if and only if $v_p(a)\ge d$.
If $p=2$, then $1+pa$ is an $m$-th power if and only if $v_p(a)\ge d+1$.
Let $f_m : 1+p\mathbb Z_p\to 1+p\mathbb Z_p$ be the $m$-th power map. This is a multiplicative group homomorphism.
Lemma Let $r\ge 1$. We have $$f_m(1+p^r\mathbb Z_p)=1+p^{r+\epsilon}\mathbb Z_p$$ with
$\epsilon=0$ if $v_p(m)=0$;
$\epsilon=1$ if $m=p>2$, or if $m=p=2$ and $r\ge 2$;
$\epsilon=2$ if $m=p=2$ and $r=1$.
Proof. That $f_m(1+p^r\mathbb Z_p)\subseteq 1+p^{r+\epsilon}\mathbb Z_p$ is clear. For the converse, fix an $a\in \mathbb Z_p$. We want to solve an equation $$(1+p^rx)^m=1+p^{r+\epsilon}a$$ with $x\in \mathbb Z_p$. In the cases 1 and 2, it is equivalent to $$ p^{rm-r-\epsilon}x^m+mp^{rm-2r-\epsilon}x^{m-1}+\cdots+mp^{-\epsilon}x-a=0. $$ The above equation modulo $p$ has degree one, hence (as polynomial in $\mathbb F_p[X]$) has a simple root in $\mathbb F_p$, so by Hensel's lemma it has a (unique) solution in $\mathbb Z_p$.
The case 3 is treated in this question.
Proof of the claim: Decompose $m=p^dq$ with $q$ prime to $p$. Then $f_m=f_{p^r}\circ f_q$. By the case 1 above, we are reduced to the case $m=p^d$. We conclude by easy induction on $d$ using cases 2 and 3.
Remark This question can certainly be solved with more involved material as formal groups and isogenies.
Let me start from @Cantlog’s response, which points out that the difficulty is in knowing when $1+pa$ is an $m$-th power. The answer depends strongly on the divisibility of $m$ by $p$. In particular, if $m$ is indivisible by $p$, then $1+pa$ is always an $m$-th power. Indeed, as @Jyrki points out, $(1+pa)^z$ has a well-defined meaning for any $p$-adic integer $z$. For, if $\{n_i\}$ is a sequence of positive integers with $p$-adic limit equal to $z$, one shows (it’s a nice exercise) that $\{(1+pa)^{n_i}\}$ is Cauchy, and you define $(1+pa)^z$ to be the limit.
When $m$ is indivisible by $p$, $1/m$ is a $p$-adic integer, and by the remarks above, you’re good to go. So the question reduces to what the $pa$’s are such that $1+pa$ is a $p$-th power. When $p>2$, you see that $(1+pb)^p=1+p^2bu$, where $u$ is a $p$-adic unit. This also is an easy consequence of the Binomial Theorem. So, as long as $a$ itself is divisible by $p$, $1+pa$ is a $p$-th power. The story is just a little more complicated for $p=2$, because the expansion of $(1+z)^2$ involves so few terms. But if $a$ is a $p$-adic unit, then $1+pa$ is definitely not a $p$-th power.
Incomplete Answer: At least for the cube root case, this - like the square root case - follows from a theorem of Katok (2007). You can also get nice approximations in both the square root case and cube root case.
For more details on the above, see (link):
Ignacio, P. S., Addawe, J., Alangui, W., & Nable, J. (2013). Computation of Square and Cube Roots of $p$-Adic Numbers via Newton-Raphson Method. Journal of Mathematics Research, 5(2), p. 31.
