Find all positive integers $m$ and $n$ such that both $3^{m}+1$ and $3^{n}+1$ are divisible by $mn$.
1. We need to find all positive integers \( m \) and \( n \) such that both \( 3^m + 1 \) and \( 3^n + 1 \) are divisible by \( mn \).

2. First, consider the case where \( m = 1 \) or \( n = 1 \). If \( m = 1 \), then \( 3^m + 1 = 3 + 1 = 4 \), and for \( 3^m + 1 \) to be divisible by \( mn \), \( n \) must be a divisor of 4. The divisors of 4 are 1, 2, and 4. Similarly, if \( n = 1 \), then \( 3^n + 1 = 4 \), and \( m \) must be a divisor of 4. Therefore, the pairs \((m, n)\) can be \((1, 1)\), \((1, 2)\), \((1, 4)\), \((2, 1)\), and \((4, 1)\).

3. Now, consider the case where \( m > 1 \) and \( n > 1 \). We need to show that no such pairs \((m, n)\) exist. 

4. Note that if \( m \) and \( n \) are both greater than 1, then one of \( m \) or \( n \) must be odd. This is because \( 3^{2k} + 1 \equiv 2 \pmod{4} \), which means \( 3^{2k} + 1 \) is not divisible by 4, and hence cannot be divisible by \( mn \) if both \( m \) and \( n \) are even.

5. Without loss of generality, assume \( m \) is odd. Since \( m \) is odd, \( 3^m + 1 \) is even. We need \( m \) to divide \( 3^m + 1 \). 

6. Let \( p \) be the smallest prime divisor of \( m \). Since \( p \) divides \( m \), it must also divide \( 3^m + 1 \). 

7. Since \( p \) is a prime divisor of \( m \), and \( \gcd(3, p) = 1 \), by Fermat's Little Theorem, we have \( 3^{p-1} \equiv 1 \pmod{p} \). 

8. Since \( p \) divides \( 3^m + 1 \), we have \( 3^m \equiv -1 \pmod{p} \). Squaring both sides, we get \( 3^{2m} \equiv 1 \pmod{p} \).

9. The order of 3 modulo \( p \), denoted \( \text{ord}_p(3) \), must divide both \( 2m \) and \( p-1 \). Since \( m \) is odd, \( \text{ord}_p(3) \) must be 2. 

10. Therefore, \( 2 \) must divide \( p-1 \), implying \( p \equiv 1 \pmod{2} \), so \( p \) is odd. 

11. Since \( p \) is the smallest prime divisor of \( m \), and \( \text{ord}_p(3) = 2 \), we have \( p-1 \) must be even, and \( p-1 \) must have a prime divisor less than \( p \), which is a contradiction because \( p \) is the smallest prime divisor of \( m \).

12. Therefore, no such \( m \) and \( n \) exist where both \( m > 1 \) and \( n > 1 \).

The final answer is \( \boxed{ (m, n) = (1, 1), (1, 2), (1, 4), (2, 1), (4, 1) } \).