Find all pairs $(a,b)$ of integers with $a>1$ and $b>1$ such that $a$ divides $b+1$ and $b$ divides $a^3-1$.
To find all pairs \((a, b)\) of integers with \(a > 1\) and \(b > 1\) such that \(a\) divides \(b+1\) and \(b\) divides \(a^3-1\), we will analyze the given conditions step by step.

1. **Condition Analysis:**
   - \(a \mid (b + 1)\) implies that \(b + 1 = ka\) for some integer \(k\).
   - \(b \mid (a^3 - 1)\) implies that \(a^3 - 1 = mb\) for some integer \(m\).

2. **Express \(b\) in terms of \(a\):**
   - From \(a \mid (b + 1)\), we have \(b + 1 = ka\).
   - Therefore, \(b = ka - 1\).

3. **Substitute \(b\) into the second condition:**
   - From \(b \mid (a^3 - 1)\), we have \(a^3 - 1 = m(ka - 1)\).
   - Substituting \(b = ka - 1\) into the equation, we get:
     \[
     a^3 - 1 = m(ka - 1)
     \]
   - Expanding and rearranging terms, we get:
     \[
     a^3 - 1 = mka - m
     \]
     \[
     a^3 - mka = m - 1
     \]
     \[
     a(a^2 - mk) = m - 1
     \]

4. **Analyze the equation \(a(a^2 - mk) = m - 1\):**
   - For \(a(a^2 - mk) = m - 1\) to hold, \(m - 1\) must be divisible by \(a\).
   - Let us consider different cases for \(m\) and \(k\).

5. **Case 1: \(m < k\):**
   - If \(m < k\), then \(\Delta = (mk)^2 - 4m + 4k < (mk + 2)^2\).
   - This implies \((m - 1)(k + 1) > -2\), hence \(\Delta < (mk + 2)^2\) if \(m \neq 1\).
   - If \(m = 0\), then \(k = a^2\) and \(b = ka - 1 = a^3 - 1\).
   - Therefore, \((a, b) = (s, s^3 - 1)\) where \(s \geq 2\) is an integer.

6. **Case 2: \(m > k\):**
   - If \(m > k\), then \(\Delta = (mk)^2 - 4m + 4k > (mk - 2)^2\).
   - This implies \((m + 1)(k - 1) > 0\), hence \(\Delta < (mk + 2)^2\) if \(k \neq 1\).
   - If \(k = 1\), then \(\Delta = (m - 2)^2\), hence \(a = m - 1\) and \(b = ka - 1 = a - 1 = m - 2\).
   - Therefore, \((a, b) = (s, s - 1)\) where \(s \geq 3\) is an integer.

Conclusion:
The pairs \((a, b)\) of integers with \(a > 1\) and \(b > 1\) that satisfy the given conditions are:
1. \((a, b) = (s, s^3 - 1)\) where \(s \geq 2\) is an integer.
2. \((a, b) = (s, s - 1)\) where \(s \geq 3\) is an integer.

The final answer is \( \boxed{ (a, b) = (s, s^3 - 1) } \) where \(s \geq 2\) and \((a, b) = (s, s - 1)\) where \(s \geq 3\).