Let $ c$ be a positive integer. The sequence $ a_1,a_2,\ldots$ is defined as follows $ a_1\equal{}c$, $ a_{n\plus{}1}\equal{}a_n^2\plus{}a_n\plus{}c^3$ for all positive integers $ n$. Find all $ c$ so that there are integers $ k\ge1$ and $ m\ge2$ so that $ a_k^2\plus{}c^3$ is the $ m$th power of some integer.
1. **Define the sequence and the problem statement:**
   Let \( c \) be a positive integer. The sequence \( a_1, a_2, \ldots \) is defined as follows:
   \[
   a_1 = c, \quad a_{n+1} = a_n^2 + a_n + c^3 \quad \text{for all positive integers } n.
   \]
   We need to find all \( c \) such that there exist integers \( k \ge 1 \) and \( m \ge 2 \) so that \( a_k^2 + c^3 \) is the \( m \)-th power of some integer.

2. **Introduce the sequence \( b_n \):**
   Define \( b_n = a_n^2 + c^3 \). We are interested in finding when \( b_n \) is a perfect power.

3. **Derive the recursion for \( b_{n+1} \):**
   \[
   b_{n+1} = a_{n+1}^2 + c^3 = (a_n^2 + a_n + c^3)^2 + c^3
   \]
   Expanding the square:
   \[
   b_{n+1} = (a_n^2 + a_n + c^3)^2 + c^3 = a_n^4 + 2a_n^3 + (2c^3 + 1)a_n^2 + 2c^3a_n + c^6 + c^3
   \]
   Notice that:
   \[
   b_{n+1} = (a_n^2 + c^3)^2 + 2a_n(a_n^2 + c^3) + a_n^2 + c^3 = b_n^2 + 2a_nb_n + a_n^2 + c^3
   \]
   Simplifying further:
   \[
   b_{n+1} = b_n(b_n + 2a_n + 1)
   \]

4. **Prove that \( b_n \) and \( b_n + 2a_n + 1 \) are coprime:**
   Assume \( p \) is a common prime divisor of \( b_n \) and \( b_n + 2a_n + 1 \). Then:
   \[
   p \mid b_n \quad \text{and} \quad p \mid (b_n + 2a_n + 1)
   \]
   This implies:
   \[
   p \mid (b_n + 2a_n + 1 - b_n) = 2a_n + 1
   \]
   Therefore:
   \[
   2a_n \equiv -1 \pmod{p}
   \]
   Substituting \( 2a_n \equiv -1 \pmod{p} \) into \( b_n = a_n^2 + c^3 \):
   \[
   0 \equiv (2a_n)^2 + 4c^3 \equiv 1 + 4c^3 \pmod{p}
   \]
   Thus:
   \[
   4c^3 \equiv -1 \pmod{p}
   \]
   Since \( 2a_n \equiv -1 \pmod{p} \), we have:
   \[
   2a_{n-1} \equiv -1 \pmod{p}
   \]
   By induction, \( p \) must divide \( 2a_1 + 1 = 2c + 1 \). Therefore:
   \[
   4c^3 \equiv -1 \pmod{p} \quad \text{and} \quad 2c \equiv -1 \pmod{p}
   \]
   This implies:
   \[
   c \equiv -1 \pmod{p} \quad \text{and} \quad p \mid 1
   \]
   This is a contradiction, so \( b_n \) and \( b_n + 2a_n + 1 \) are coprime.

5. **Induction on \( b_n \) being an \( m \)-th power:**
   If \( b_{n+1} \) is an \( m \)-th power, then so is \( b_n \). By induction, \( b_1 = c^3 + c^2 \) must also be an \( m \)-th power.

6. **Factorize \( b_1 \):**
   \[
   b_1 = c^3 + c^2 = c^2(c + 1)
   \]
   Since \( c^2 \) and \( c + 1 \) are coprime, both must be \( m \)-th powers. Therefore, \( c \) must be a perfect square, say \( c = k^2 \).

7. **Solve for \( c \):**
   If \( c = k^2 \), then:
   \[
   c + 1 = k^2 + 1
   \]
   For \( c + 1 \) to be a perfect square, let \( c + 1 = \ell^2 \). Thus:
   \[
   c = \ell^2 - 1
   \]
   Therefore, all possible \( c \) are of the form \( \ell^2 - 1 \) for any integer \( \ell \geq 2 \).

The final answer is \( \boxed{ c = \ell^2 - 1 } \) for any integer \( \ell \geq 2 \).