We are tasked with finding all positive integers $a, n \ge 1$ such that for all primes $p$ dividing $a^n - 1$, there exists a positive integer $m < n$ such that $p \mid a^m - 1$.

By Zsigmondy's theorem, for any $a > 1$ and $n > 1$, there exists a primitive prime divisor of $a^n - 1$ except for the cases $(a, n) = (2, 6)$ and $(a, n) = (2^k - 1, 2)$.

1. **Case $(a, n) = (2, 6)$:**
- For $a = 2$ and $n = 6$, we have $2^6 - 1 = 63$.
- The prime divisors of 63 are 3 and 7.
- We need to check if these primes divide $2^m - 1$ for some $m < 6$.
- The values of $2^m - 1$ for $m < 6$ are: $1, 3, 7, 15, 31$.
- Both 3 and 7 appear in this list, so this case holds.

2. **Case $(a, n) = (2^k - 1, 2)$:**
- For $a = 2^k - 1$ and $n = 2$, we have $(2^k - 1)^2 - 1 = (2^k - 1)(2^k + 1)$.
- Any prime divisor of $(2^k - 1)^2 - 1$ must divide either $2^k - 1$ or $2^k + 1$.
- The only possible $m$ is $m = 1$, which gives us $2^k - 1 - 1 = 2^k - 2 = 2(2^{k-1} - 1)$.
- Thus, all prime divisors of $(2^k - 1)^2 - 1$ divide $2(2^{k-1} - 1)$, satisfying the condition.

3. **Case $a = 1$:**
- For $a = 1$, $1^n - 1 = 0$ for any $n$, which trivially satisfies the condition.

Combining these results, the complete solution set is:
$$(a, n) = (2, 6), (2^k - 1, 2), (1, n) \text{ for any } n \ge 1.$$

The answer is: \boxed{(2, 6), (2^k - 1, 2), (1, n) \text{ for any } n \ge 1}.