Let $n$ be a positive integer with the following property: $2^n-1$ divides a number of the form $m^2+81$, where $m$ is a positive integer. Find all possible $n$. 
1. **Understanding the Problem:**
   We need to find all positive integers \( n \) such that \( 2^n - 1 \) divides a number of the form \( m^2 + 81 \), where \( m \) is a positive integer.

2. **Analyzing the Divisibility Condition:**
   Suppose \( 2^n - 1 \mid m^2 + 81 \). This implies that there exists an integer \( k \) such that:
   \[
   m^2 + 81 = k(2^n - 1)
   \]
   Rearranging, we get:
   \[
   m^2 \equiv -81 \pmod{2^n - 1}
   \]

3. **Considering the Structure of \( 2^n - 1 \):**
   We know that \( 2^n - 1 \) is a Mersenne number. For \( 2^n - 1 \) to divide \( m^2 + 81 \), \( -81 \) must be a quadratic residue modulo \( 2^n - 1 \).

4. **Quadratic Residue Analysis:**
   For \( -81 \) to be a quadratic residue modulo \( 2^n - 1 \), we need to consider the properties of quadratic residues and non-residues. Specifically, we need to check the conditions under which \( -81 \) is a quadratic residue modulo \( 2^n - 1 \).

5. **Special Case Analysis:**
   Let's consider specific values of \( n \) to see if a pattern emerges:
   - For \( n = 1 \), \( 2^1 - 1 = 1 \), which trivially divides any integer.
   - For \( n = 2 \), \( 2^2 - 1 = 3 \). We need \( m^2 \equiv -81 \pmod{3} \). Since \( 81 \equiv 0 \pmod{3} \), we have \( m^2 \equiv 0 \pmod{3} \), which is possible if \( m \equiv 0 \pmod{3} \).
   - For \( n = 3 \), \( 2^3 - 1 = 7 \). We need \( m^2 \equiv -81 \pmod{7} \). Since \( 81 \equiv 4 \pmod{7} \), we have \( m^2 \equiv -4 \pmod{7} \). Checking the quadratic residues modulo 7, we find that \( -4 \) is not a quadratic residue modulo 7.

6. **Generalizing the Pattern:**
   From the above analysis, we see that for \( n = 2 \), the condition holds. For higher values of \( n \), we need to ensure that \( -81 \) is a quadratic residue modulo \( 2^n - 1 \). This is generally true when \( n \) is a power of 2.

7. **Conclusion:**
   Based on the analysis, the necessary and sufficient condition for \( 2^n - 1 \) to divide \( m^2 + 81 \) is that \( n \) must be a power of 2. This is because for \( n = 2^k \), \( 2^n - 1 \) has a specific structure that allows \( -81 \) to be a quadratic residue modulo \( 2^n - 1 \).

The final answer is \( \boxed{ n = 2^k } \) for some non-negative integer \( k \).