Let $a,b,m,n$ integers greater than 1. If $a^n-1$ and $b^m+1$ are both primes, give as much info as possible on $a,b,m,n$.
1. **Analyzing \( b^m + 1 \) being prime:**

   We start by considering the expression \( b^m + 1 \) and assume it is a prime number. We will show that \( m \) must be a power of 2.

   Suppose \( m = 2^k l \) where \( k \in \mathbb{N} \) and \( l > 1 \) is odd. Let \( c = b^{2^k} \). Then we can rewrite \( b^m + 1 \) as:
   \[
   b^m + 1 = c^l + 1
   \]
   Since \( c > 2 \), by **Zsigmondy's theorem**, there exists a prime \( p \) such that \( p \) divides \( c^l + 1 \) and \( p \) does not divide \( c + 1 \). This implies that there is another prime \( q \neq p \) such that \( q \) divides \( c + 1 \).

   Now, consider the factorization:
   \[
   c^l + 1 = (c + 1)(c^{l-1} - c^{l-2} + \dots + 1)
   \]
   Since \( p \) divides \( c^l + 1 \) and \( q \) divides \( c + 1 \), both \( p \) and \( q \) must divide \( c^l + 1 \). This leads to a contradiction because \( c^l + 1 \) is assumed to be a prime number, which cannot have more than one distinct prime factor. Therefore, our assumption that \( l > 1 \) must be false, and we conclude that \( l = 1 \). Hence, \( m \) must be a power of 2:
   \[
   m = 2^k \quad \text{for some} \quad k \in \mathbb{N}
   \]

2. **Analyzing \( a^n - 1 \) being prime:**

   Next, we consider the expression \( a^n - 1 \) and assume it is a prime number. We will show that \( n \) must be 2.

   Suppose \( n \) is an integer greater than 1. If \( n \) is not 2, then \( n \) must be composite, say \( n = rs \) where \( r, s > 1 \). Let \( d = a^r \). Then we can rewrite \( a^n - 1 \) as:
   \[
   a^n - 1 = d^s - 1
   \]
   Using the factorization of \( d^s - 1 \):
   \[
   d^s - 1 = (d - 1)(d^{s-1} + d^{s-2} + \dots + 1)
   \]
   Since \( d > 2 \), both \( d - 1 \) and \( d^{s-1} + d^{s-2} + \dots + 1 \) are greater than 1, implying that \( a^n - 1 \) is not a prime number. This leads to a contradiction. Therefore, \( n \) must be 2:
   \[
   n = 2
   \]

Conclusion:
- \( m \) must be a power of 2: \( m = 2^k \) for some \( k \in \mathbb{N} \).
- \( n \) must be 2: \( n = 2 \).

The final answer is \( \boxed{ m = 2^k } \) for some \( k \in \mathbb{N} \) and \( n = 2 \).