Find all pairs of integers $ \left(m,n\right) $ such that $ \left(m,n+1\right) = 1 $ and $$ \sum\limits_{k=1}^{n}{\frac{m^{k+1}}{k+1}\binom{n}{k}} \in \mathbb{N} $$
1. We start with the given expression:
   \[
   \sum_{k=1}^{n} \frac{m^{k+1}}{k+1} \binom{n}{k}
   \]
   We need to find pairs of integers \((m, n)\) such that \((m, n+1) = 1\) and the above sum is an integer.

2. Using a known result from calculus, we can simplify the given sum:
   \[
   \sum_{k=1}^{n} \frac{m^{k+1}}{k+1} \binom{n}{k} = \frac{(m+1)^{n+1} + 1}{n+1} - m
   \]
   This result can be derived using the binomial theorem and properties of binomial coefficients.

3. We need the expression:
   \[
   \frac{(m+1)^{n+1} + 1}{n+1} - m
   \]
   to be an integer. This implies:
   \[
   \frac{(m+1)^{n+1} + 1}{n+1} \in \mathbb{N}
   \]
   Therefore, \((m+1)^{n+1} + 1\) must be divisible by \(n+1\).

4. Let \(d = n+1\). Then, we need:
   \[
   d \mid (m+1)^{d} + 1
   \]
   and \((m, d) = 1\).

5. Consider the case when \(d\) is odd. Let \(m+1 = kd - 1\) for some integer \(k\). Then:
   \[
   (m+1)^{d} = (kd - 1)^{d}
   \]
   Using the binomial expansion:
   \[
   (kd - 1)^{d} = \sum_{i=0}^{d} \binom{d}{i} (kd)^{d-i} (-1)^{i}
   \]
   The term \(\binom{d}{0} (kd)^{d} (-1)^{0} = (kd)^{d}\) is clearly divisible by \(d\). All other terms in the expansion are also divisible by \(d\) because they contain \(kd\) as a factor. Therefore:
   \[
   (kd - 1)^{d} \equiv (-1)^{d} \pmod{d}
   \]
   Since \(d\) is odd, \((-1)^{d} = -1\). Thus:
   \[
   (kd - 1)^{d} \equiv -1 \pmod{d}
   \]
   Hence:
   \[
   (m+1)^{d} + 1 \equiv 0 \pmod{d}
   \]
   which means \(d \mid (m+1)^{d} + 1\).

6. Therefore, for \(d = n+1\) odd, \(m+1 = kd - 1\) works. This gives:
   \[
   m = kd - 2 = k(n+1) - 2
   \]

7. We need to check the condition \((m, n+1) = 1\). Since \(m = k(n+1) - 2\), we have:
   \[
   (k(n+1) - 2, n+1) = (n-1, n+1)
   \]
   Since \(n\) is even, \((n-1, n+1) = 1\).

8. Therefore, the set of solutions is:
   \[
   (m, n) = (k(n+1) - 2, n)
   \]
   for all even \(n\) and all \(k \in \mathbb{N}\).

The final answer is \( \boxed{ (m, n) = (k(n+1) - 2, n) } \) for all even \(n\) and all \(k \in \mathbb{N}\).