Consider the polynomial $f(x)=x^{2m}-p^{2n}=0$ over the $p$-adic field $\mathbb Q_p$. Note that $m,n$ are positive integers and $\gcd(m,n)=1$. I want to study the field extension by the zeros of $f(x)$.
We can rewrite $f(x)=(x^m-p^n)(x^m+p^n)=f_1(x) f_2(x)$, where $f_1(x)$ and $f_2(x)$ are Eisenstein polynomials and of course irreducible since $\gcd(m,n)=1$.
Clearly, zeros of $f_1(x)$ give a degree $m$ extension of $\mathbb Q_p$, which is totally ramified, and the same is true for the zeros of $f_2(x)$.
But the compositum of totally ramified extensions may not be compositum.
What is the degree of extension? When is it totally ramified?
In both cases of $f_1(x)$ and $f_2(x)$, the unifomizer is $p$. Thus, I think the compositum will be totally ramified. I found some discussion here
Let's work in a fixed algebraic closure of $\mathbb Q_p$. You say (in a comment) you want to adjoin all roots of $f(x)$, i.e. you are looking at the splitting field of that polynomial. Let $\alpha$ be any one such root.
Then it's basic theory of cyclic field extensions (this is the "easy direction" of Kummer theory) that this splitting field can be seen as the compositum of, on the one hand, $\mathbb Q_p(\alpha)$, and on the other hand, $K_{2m} := \mathbb Q_p(\mu_{2m})$, the cyclotomic extension by the $2m$-th roots of unity.
Now for sure, the first field $\mathbb Q_p(\alpha)$ is totally ramified: As discussed in comments, the polynomial $f$ and/or its factors $f_1, f_2$ are technically only Eisenstein for $n=1$, but the proof idea for this generalizes immediately. Alternatively one could see this with the Newton polygon (which has only one slope), or even more basic by noticing that i) the extension has degree $\le m$ and ii) because $m,n$ are coprime, there exist $k, \ell$ such that $v_p(\frac{\alpha^k}{p^\ell}) = \frac{1}{m}$ i.e. the value group of $\mathbb Q_p$ has index at least $m$ in the value group of $\mathbb Q_p(\alpha)$.
But that second field, the cyclotomic $K_{2m}$, is in general not totally ramified: By well-known facts about roots of unity in $p$-adics, this field contains a proper unramified extension unless $2m$ is of the form $p^r \cdot d$, with $d$ a divisor of $p-1$, and $r \in\{0,1,2,\dots\}$. (Here and from now on, I assume $p$ is odd, and leave the case $p=2$ to you.)
So $2m$ being "a power of $p$, times a divisor of $p-1$" is a necessary criterion for our composite splitting field to be totally ramified.
Now in case $2m | (p-1)$ (i.e. $r=0$ i.e. $K_{2m} = \mathbb Q_p$), of course the compositum is just the totally ramified $\mathbb Q_p(\alpha)$, so it is totally ramified. That is, $2m$ being a divisor of $p-1$ happens to be sufficient for what you want.
But in the remaining cases $2m = p^r \cdot d$, $r \ge 1$, I do not see right away whether the compositum of our two fields (albeit each of them is totally ramified) is also totally ramified. That needs a closer look, and unlike anything so far, might even depend on whether $\alpha$ is a root of $f_1$ or $f_2$. (I described in comments the slight distinction between roots of the one and the other. Note also that e.g. in the case $m=p-1$ you brought up, the splitting fields of each of the factors $f_1, f_2$, respectively, are totally ramified by slightly altered arguments -- now $m | (p-1)$ is enough --, but their compositum, i.e. the splitting field of $f$, is not, as per above.)