 Let $m$ and $n$ be two positive integers, $m, n \ge 2$. Solve in the set of the positive integers the equation $x^n + y^n = 3^m$.
1. **Applying Zsigmondy's Theorem:**
   Zsigmondy's theorem states that for any two coprime integers \(a\) and \(b\) (with \(a > b > 0\)) and any integer \(n > 1\), there exists a prime number \(p\) (called a primitive prime divisor) that divides \(a^n - b^n\) but does not divide \(a^k - b^k\) for any \(1 \le k < n\), with some exceptions. 

   In our problem, we have \(x^n + y^n = 3^m\). Since \(3\) is the only prime divisor of \(3^m\), by Zsigmondy's theorem, \(3\) must be a primitive prime divisor of \(x^n + y^n\). This implies that \(3\) does not divide \(x + y\) unless \((x, y, n)\) is one of the exceptions.

2. **Considering the exceptions:**
   The exceptions to Zsigmondy's theorem are:
   - \(a = 2, b = 1, n = 3\)
   - \(a = 1, b = -1, n = 2\)
   - \(a = 1, b = 1, n = 2\)

   In our case, the relevant exception is \((x, y, n) = (2, 1, 3)\) or \((1, 2, 3)\).

3. **Checking the exception:**
   For \((x, y, n) = (2, 1, 3)\):
   \[
   2^3 + 1^3 = 8 + 1 = 9 = 3^2
   \]
   Thus, \(m = 2\).

   Similarly, for \((x, y, n) = (1, 2, 3)\):
   \[
   1^3 + 2^3 = 1 + 8 = 9 = 3^2
   \]
   Thus, \(m = 2\).

4. **Considering the case \(3 \not| x + y\):**
   If \(3 \not| x + y\), then \(3\) must divide both \(x\) and \(y\) (since \(3\) is the only prime divisor of \(x^n + y^n\)). However, if \(3\) divides both \(x\) and \(y\), then \(x = 3k\) and \(y = 3l\) for some integers \(k\) and \(l\). This would imply:
   \[
   (3k)^n + (3l)^n = 3^m
   \]
   \[
   3^n (k^n + l^n) = 3^m
   \]
   \[
   k^n + l^n = 3^{m-n}
   \]
   Since \(k\) and \(l\) are integers, \(k^n + l^n\) must be a power of \(3\). However, this is not possible unless \(k = 1\) and \(l = 0\) or vice versa, which contradicts the assumption that \(x\) and \(y\) are both non-zero positive integers.

5. **Considering \(n\) is even:**
   If \(n\) is even, then:
   \[
   x^n + y^n \equiv 1 + 1 \equiv 2 \pmod{3}
   \]
   This is a contradiction since \(x^n + y^n\) must be divisible by \(3\).

6. **Considering \(n\) is odd:**
   If \(n\) is odd, then:
   \[
   x^n + y^n \equiv x + y \not\equiv 0 \pmod{3}
   \]
   This is also a contradiction since \(x^n + y^n\) must be divisible by \(3\).

Thus, the only solutions are \((x, y, n, m) = (2, 1, 3, 2)\) and \((1, 2, 3, 2)\).

The final answer is \( \boxed{ (x, y, n, m) = (2, 1, 3, 2) } \) and \((1, 2, 3, 2)\).