Upon dividing the given polynomial by $m^{2k}$ and replacing $x$ with $mx+m$, it assumes the simpler form
\begin{equation}
s+(x+1)^{2k+1}-x^{2k+1}
\end{equation}
for some $s$.

Generically its Galois group is the wreath product $C_2^k\rtimes S_k$. This follow from Hilbert's irreducibility theorem together with the fact that its Galois group over $\mathbb C(s)$, where we consider $s$ as a transcendental over the complex numbers, is $C_2^k\rtimes S_k$, so in general the roots cannot be expressed in radicals once $k\ge5$.

Write $n=2k$, set $f(x)=(x+1)^{n+1}-x^{n+1}$ and consider the branched cover $a\mapsto f(a)$. First note that $f(x-1/2)$ is an even function, so the monodromy group $G$ of this cover is a subgroup of $C_2^k\rtimes S_k$.

The critical points of the cover are the roots of $f'(x)$. Let $\gamma$ be such a root, so $(\gamma+1)^n-\gamma^n=0$ and therefore $f(\gamma)=\gamma^n$. The possibilities for $\gamma$ are $\gamma=\frac{1}{\zeta-1}$, where $\zeta$ is an $n$-th root of unity different from $1$. Note that $-\gamma$ corresponds to $1/\zeta$. Looking at absolute values we see that each $f(\gamma_1)=f(\gamma_2)$ if and only if $\gamma_1$ and $\gamma_2$ are complex conjugate. Note that $\gamma=-1/2$ corresponds to $\zeta=-1$. (Recall that $n$ is even.) So we see that the finite generators of the monodromy group of the cover is one transposition and $(n-2)/2$ double transpositions. The double transpositions induce transpositions on the blocks, so the action on the blocks is $S_k$. The transposition fixes all blocks and flips the two elements in one block. This together yields the claimed structure of the group.