Find all integers $k\ge 2$ such that for all integers $n\ge 2$,  $n$ does not divide the greatest odd divisor of $k^n+1$.
1. **Consider the problem statement**: We need to find all integers \( k \ge 2 \) such that for all integers \( n \ge 2 \), \( n \) does not divide the greatest odd divisor of \( k^n + 1 \).

2. **Let \( n = p \) where \( p \) is a prime number**: This simplifies the problem since we can use properties of prime numbers to analyze the divisibility conditions.

3. **Analyze the expression \( k^p + 1 \) modulo \( p \)**:
   \[
   k^p + 1 \equiv k + 1 \pmod{p}
   \]
   This follows from Fermat's Little Theorem, which states \( k^p \equiv k \pmod{p} \) for any integer \( k \) and prime \( p \).

4. **Conclude that \( p \) does not divide \( k + 1 \)**:
   \[
   p \mid k + 1 \implies p \mid k^p + 1
   \]
   Since \( p \) is any prime, \( k + 1 \) must not be divisible by any prime \( p \). Therefore, \( k + 1 \) must be a power of 2, i.e., \( k + 1 = 2^l \) for some integer \( l \ge 1 \).

5. **Express \( k \) in terms of \( l \)**:
   \[
   k = 2^l - 1
   \]

6. **Verify the condition for \( k = 2^l - 1 \)**:
   - Let \( k = 2^l - 1 \).
   - Consider \( n \) to be an odd integer.
   - For an odd prime \( p \), denote \( q(p) \) as the minimal odd \( m \) such that \( p \mid k^m + 1 \). If no such \( m \) exists, let \( q(p) = \infty \).

7. **Analyze the divisibility condition**:
   \[
   n \mid k^n + 1 \implies \forall p \mid n, \ q(p) \mid n
   \]
   - Let \( p \mid n \) be a prime divisor such that \( q(p) \) is minimal.
   - If \( p \) is minimal, then \( q(p) = 1 \) or \( p \mid k + 1 = 2^l \), which is a contradiction since \( p \) is an odd prime.
   - If \( p \) is not minimal, then for \( p_0 < p \) where \( p_0 \mid n \), we get \( q(p_0) = \infty \).

8. **Conclude that \( k = 2^l - 1 \) works for all \( l \ge 2 \)**:
   - This satisfies the condition that \( n \) does not divide the greatest odd divisor of \( k^n + 1 \) for all \( n \ge 2 \).

The final answer is \( \boxed{ k = 2^l - 1 } \) for \( l \ge 2 \).