Solve the equation $m^{gcd(m,n)} = n^{lcm(m,n)}$ in positive integers, where gcd($m, n$) – greatest common
divisor of $m,n$, and lcm($m, n$) – least common multiple of $m,n$.
1. **Express \( m \) and \( n \) in terms of their greatest common divisor (gcd):**
   Let \( d = \gcd(m, n) \). Then we can write \( m = d \cdot a \) and \( n = d \cdot b \), where \( \gcd(a, b) = 1 \).

2. **Rewrite the given equation using these expressions:**
   The given equation is \( m^{\gcd(m, n)} = n^{\mathrm{lcm}(m, n)} \).
   Substituting \( m = d \cdot a \) and \( n = d \cdot b \), we get:
   \[
   (d \cdot a)^{d} = (d \cdot b)^{\mathrm{lcm}(d \cdot a, d \cdot b)}
   \]

3. **Simplify the exponents:**
   Since \( \gcd(m, n) = d \), we have:
   \[
   (d \cdot a)^{d} = (d \cdot b)^{\mathrm{lcm}(d \cdot a, d \cdot b)}
   \]
   The least common multiple (lcm) of \( d \cdot a \) and \( d \cdot b \) is:
   \[
   \mathrm{lcm}(d \cdot a, d \cdot b) = d \cdot \mathrm{lcm}(a, b)
   \]
   Since \( \gcd(a, b) = 1 \), we have \( \mathrm{lcm}(a, b) = a \cdot b \). Therefore:
   \[
   \mathrm{lcm}(d \cdot a, d \cdot b) = d \cdot a \cdot b
   \]

4. **Substitute back into the equation:**
   \[
   (d \cdot a)^{d} = (d \cdot b)^{d \cdot a \cdot b}
   \]
   Simplifying the left-hand side:
   \[
   d^d \cdot a^d = d^{d \cdot a \cdot b} \cdot b^{d \cdot a \cdot b}
   \]

5. **Equate the powers of \( d \):**
   Since \( d^d \cdot a^d = d^{d \cdot a \cdot b} \cdot b^{d \cdot a \cdot b} \), we can equate the powers of \( d \):
   \[
   d^d = d^{d \cdot a \cdot b}
   \]
   This implies:
   \[
   d = d \cdot a \cdot b
   \]
   Since \( d \neq 0 \), we can divide both sides by \( d \):
   \[
   1 = a \cdot b
   \]

6. **Solve for \( a \) and \( b \):**
   The only positive integer solutions for \( a \) and \( b \) that satisfy \( a \cdot b = 1 \) are \( a = 1 \) and \( b = 1 \).

7. **Substitute back to find \( m \) and \( n \):**
   Since \( a = 1 \) and \( b = 1 \), we have:
   \[
   m = d \cdot a = d \cdot 1 = d
   \]
   \[
   n = d \cdot b = d \cdot 1 = d
   \]
   Since \( \gcd(m, n) = d \), we have \( d = 1 \).

8. **Conclusion:**
   Therefore, \( m = 1 \) and \( n = 1 \) are the only solutions.

The final answer is \( \boxed{ (m, n) = (1, 1) } \).