Find all positive integers $k$ such that there exists a positive integer $n$, for which $2^n + 11$ is divisible by $2^k - 1$.
To solve the problem, we need to find all positive integers \( k \) such that there exists a positive integer \( n \) for which \( 2^n + 11 \) is divisible by \( 2^k - 1 \). 

1. **Understand the divisibility condition:**
   \[
   2^n + 11 \equiv 0 \pmod{2^k - 1}
   \]
   This implies:
   \[
   2^n \equiv -11 \pmod{2^k - 1}
   \]

2. **Simplify the congruence:**
   Since \( 2^k - 1 \) is a divisor of \( 2^n + 11 \), we can write:
   \[
   2^n \equiv -11 \pmod{2^k - 1}
   \]
   This means that \( 2^n \) and \(-11\) leave the same remainder when divided by \( 2^k - 1 \).

3. **Consider the properties of \( 2^k - 1 \):**
   Note that \( 2^k - 1 \) is a Mersenne number, which is always odd. Therefore, \(-11\) modulo \( 2^k - 1 \) is equivalent to \( 2^k - 1 - 11 \):
   \[
   -11 \equiv 2^k - 12 \pmod{2^k - 1}
   \]
   Thus, we have:
   \[
   2^n \equiv 2^k - 12 \pmod{2^k - 1}
   \]

4. **Analyze the congruence:**
   For the congruence \( 2^n \equiv 2^k - 12 \pmod{2^k - 1} \) to hold, \( 2^k - 12 \) must be a power of 2. Let \( 2^k - 12 = 2^m \) for some integer \( m \). This gives:
   \[
   2^k = 2^m + 12
   \]

5. **Solve for \( k \):**
   Since \( 2^k \) and \( 2^m \) are both powers of 2, we need to find \( k \) such that \( 2^k = 2^m + 12 \). This equation can be solved by checking small values of \( k \) and \( m \):
   - For \( k = 4 \):
     \[
     2^4 = 16 \quad \text{and} \quad 2^m + 12 = 16 \implies 2^m = 4 \implies m = 2
     \]
     This works since \( 2^4 = 16 \) and \( 2^2 + 12 = 16 \).

6. **Verify the solution:**
   We need to check if there are any other values of \( k \) that satisfy the equation \( 2^k = 2^m + 12 \). For \( k > 4 \), the difference between \( 2^k \) and \( 2^m \) becomes larger than 12, making it impossible for \( 2^k - 12 \) to be a power of 2.

Therefore, the only solution is \( k = 4 \).

The final answer is \( \boxed{ k = 4 } \).