There exists no such $(m, n)$.
Suppose $(m, n)$ is a solution. By the fact that $(n + 1) \binom{n}{k} = (k + 1) \binom{n+1}{k+1}$,
we have
$$\begin{aligned} \sum_{k=1}^{n} \frac{m^{k+1}}{k+1} \binom{n}{k} &= \sum_{k=1}^{n} \frac{m^{k+1}}{n+1} \binom{n+1}{k+1} = \frac{1}{n+1} \sum_{l=0}^{n+1} m^l \binom{n+1}{l} \\ &= \frac{1}{n+1} ((1+m)^{n+1} - 1). \end{aligned}$$
Since this is an integer, we must have
$$(1+m)^{n+1} \equiv 1 \pmod{n+1}.$$
This ensures that when $m$ is odd, $(n+1)$ must be odd. On the other hand, since $\gcd(m, n + 1) = 1$, if $m$ is even, $(n + 1)$ would also be odd. So, in any case, $(n + 1)$ is odd.
Let $p$ be the smallest prime divisor of $(n + 1)$, then $p$ is odd, and
$$(1+m)^{n+1} \equiv 1 \pmod{p}.$$
However, by Fermat's little theorem, we have $(1+m)^{p-1} \equiv 1 \pmod{p}$.
This means that
$$(1+m)^{\gcd(n+1,p-1)} \equiv 1 \pmod{p}.$$
However, since $p$ is the smallest prime divisor of $n+1$, $\gcd(n+1, p-1) = 1$, so we get $(1+m) \equiv 1 \pmod{p}$, which means that $p|m \Rightarrow p|\gcd(m, n+1)$.
Contradiction!