To solve the problem, we need to find all pairs of positive integers $(m, n)$ such that $9^{|m-n|} + 3^{|m-n|} + 1$ is divisible by both $m$ and $n$ simultaneously.

Let's denote $d = |m-n|$. The expression becomes $f(d) = 9^d + 3^d + 1$.

### Step-by-step Analysis:

1. **Case $d = 0$:**
- If $d = 0$, then $m = n$.
- In this case, $f(d) = 9^0 + 3^0 + 1 = 1 + 1 + 1 = 3$.
- Therefore, $f(0) = 3$, and we need $3$ to be divisible by $m$ and $n$, which implies $m = n = 3$.

2. **Case $d > 0$:**
- If $d > 0$, we consider the smallest values of $d$ and explore possibilities:
- For $d = 1$:
$$f(1) = 9^1 + 3^1 + 1 = 9 + 3 + 1 = 13$$
- Here, we require $13$ to be divisible by both $m$ and $n$. This implies $m, n \geq 13$, but since no positive integer smaller than $13$ divides $13$ except $1$ itself, this leads us back to $m = n = 1$.

- For larger $d$ values, note that $f(d)$ grows significantly because both $9^d$ and $3^d$ grow exponentially. Thus, it is less likely for $f(d)$ to be divisible by smaller numbers simultaneously unless specific conditions hold.

3. Conclusion:
- We verify that for $m = n = 3$ and $m = n = 1$, the function $f(d)$ is divisible by both numbers:
- For $m = n = 3$, $f(0) = 3$ is satisfied.
- For $m = n = 1$, $f(0) = 3$ divided by $1$ gives no issues.

Therefore, the solutions to the problem, considering symmetry and the nature of the function, are:
$$\boxed{(1, 1) \text{ and } (3, 3)}$$