Find all pairs $(m, n)$ of positive integers, for which number $2^n - 13^m$ is a cube of a positive integer.

[i]Proposed by Oleksiy Masalitin[/i]
1. **Initial Setup:**
   Let \(2^n - 13^m = k^3\) for some \(m, n, k \in \mathbb{N}\). We need to find all pairs \((m, n)\) such that this equation holds.

2. **Modulo Analysis:**
   We start by considering \(2^n \mod 13\). The powers of 2 modulo 13 cycle every 12 steps:
   \[
   2^1 \equiv 2, \quad 2^2 \equiv 4, \quad 2^3 \equiv 8, \quad 2^4 \equiv 3, \quad 2^5 \equiv 6, \quad 2^6 \equiv 12, \quad 2^7 \equiv 11, \quad 2^8 \equiv 9, \quad 2^9 \equiv 5, \quad 2^{10} \equiv 10, \quad 2^{11} \equiv 7, \quad 2^{12} \equiv 1 \pmod{13}.
   \]
   Since \(2^n \equiv k^3 + 13^m \pmod{13}\), and \(13^m \equiv 0 \pmod{13}\), we have:
   \[
   2^n \equiv k^3 \pmod{13}.
   \]
   The cubic residues modulo 13 are \(0, 1, 8, 12\). Therefore, \(2^n \equiv 0, 1, 8, 12 \pmod{13}\). This implies \(n\) must be a multiple of 3 because only \(2^3 \equiv 8 \pmod{13}\) fits the cubic residue condition.

3. **Setting \(n = 3t\):**
   Let \(n = 3t\). Then we have:
   \[
   2^{3t} - 13^m = k^3.
   \]
   We can factorize \(2^{3t} - k^3\) as:
   \[
   (2^t - k)(2^{2t} + 2^t k + k^2) = 13^m.
   \]
   Let \(\alpha\) and \(\beta\) be non-negative integers such that \(\alpha + \beta = m\). We have two cases:
   \[
   2^t - k = 13^\alpha \quad \text{and} \quad 2^{2t} + 2^t k + k^2 = 13^\beta.
   \]

4. **Case Analysis:**
   - If \(\alpha > 0\), then \(2^t \equiv k \pmod{13}\). Substituting \(k = 2^t - 13^\alpha\) into the second equation:
     \[
     2^{2t} + 2^t (2^t - 13^\alpha) + (2^t - 13^\alpha)^2 = 13^\beta.
     \]
     Simplifying, we get:
     \[
     2^{2t} + 2^{2t} - 2^t \cdot 13^\alpha + 2^{2t} - 2 \cdot 2^t \cdot 13^\alpha + 13^{2\alpha} = 13^\beta.
     \]
     This simplifies to:
     \[
     3 \cdot 2^{2t} - 3 \cdot 2^t \cdot 13^\alpha + 13^{2\alpha} = 13^\beta.
     \]
     Since \(2^t \equiv k \pmod{13}\), we have \(2^{2t} \equiv k^2 \pmod{13}\). This leads to a contradiction because \(2^{2t} + 2^t k + k^2 \equiv 3 \cdot 2^{2t} \not\equiv 0 \pmod{13}\).

   - Therefore, \(\alpha = 0\) and \(2^t = k + 1\). Substituting \(k = 2^t - 1\) into the equation:
     \[
     3 \cdot 2^t (2^t - 1) = 13^m - 1.
     \]
     Simplifying, we get:
     \[
     3 \cdot 2^t (2^t - 1) = 13^m - 1.
     \]

5. **Finding Solutions:**
   - For \(t = 3\), we have:
     \[
     3 \cdot 2^3 (2^3 - 1) = 3 \cdot 8 \cdot 7 = 168.
     \]
     We need \(13^m - 1 = 168\), so \(13^m = 169\), which gives \(m = 2\). Thus, \(t = 3\) and \(m = 2\) is a solution, yielding \((m, n, k) = (2, 9, 7)\).

   - For \(t \ge 4\), using the lifting the exponent lemma or direct factorization, we have:
     \[
     v_2(13^m - 1) = v_2(m) + 2 = t.
     \]
     Hence, \(v_2(m) = t - 2\). Therefore, \(2^{t-2} \mid m\). Moreover, \(2^{t-2} \ge t\) for \(t \ge 4\), leading to:
     \[
     2^{2t+2} > 3 \cdot 2^{2t} = 3 \cdot 2^t - 1 + 13^m > 13^t > 2^{3t} \Rightarrow t < 2,
     \]
     which is a contradiction.

Conclusion:
The only solution is \((m, n) = (2, 9)\).

The final answer is \( \boxed{ (2, 9) } \).