Find all the pairs of positive integers $ (a,b)$ such that $ a^2 \plus{} b \minus{} 1$ is a power of prime number $ ; a^2 \plus{} b \plus{} 1$ can divide $ b^2 \minus{} a^3 \minus{} 1,$ but it can't divide $ (a \plus{} b \minus{} 1)^2.$
1. **Given Conditions and Initial Setup:**
   - We need to find pairs of positive integers \((a, b)\) such that:
     1. \(a^2 + b + 1\) is a power of a prime number.
     2. \(a^2 + b + 1\) divides \(b^2 - a^3 - 1\).
     3. \(a^2 + b + 1\) does not divide \((a + b - 1)^2\).

2. **Expressing \(a^2 + b + 1\) as a Power of a Prime:**
   - Let \(a^2 + b + 1 = p^k\), where \(p\) is a prime and \(k\) is a positive integer.

3. **Divisibility Condition:**
   - We need \(p^k\) to divide \(b^2 - a^3 - 1\).
   - Substitute \(b = p^k - a^2 - 1\) into \(b^2 - a^3 - 1\):
     \[
     b^2 - a^3 - 1 = (p^k - a^2 - 1)^2 - a^3 - 1
     \]
     \[
     = (p^k - a^2 - 1)^2 - a^3 - 1
     \]
     \[
     = p^{2k} - 2p^k a^2 + a^4 - 2p^k + 2a^2 + 1 - a^3 - 1
     \]
     \[
     = p^{2k} - 2p^k a^2 - 2p^k + a^4 + 2a^2 - a^3
     \]

4. **Non-Divisibility Condition:**
   - We need \(p^k\) to not divide \((a + b - 1)^2\).
   - Substitute \(b = p^k - a^2 - 1\) into \((a + b - 1)^2\):
     \[
     (a + b - 1)^2 = (a + p^k - a^2 - 2)^2
     \]
     \[
     = (p^k - a^2 + a - 2)^2
     \]

5. **Valuation Analysis:**
   - Let \(v_p(a) = d\). Since \(a^2 < p^k\), we have \(d < \frac{k}{2}\).
   - Since \(p^k \mid a^2(a^2 - a + 2)\), we have \(v_p(a^2 - a + 2) \geq k - 2d\).
   - But \(p^k \nmid (a^2 - a + 2)\), so \(v_p(a^2 - a + 2) < \frac{k}{2}\).
   - Thus, \(\frac{k}{2} > k - 2d\), which gives \(4d > k > 0\), hence \(p \mid a\).

6. **Case Analysis:**
   - **Case 1: \(p\) is odd.**
     - If \(p\) is odd, then \(p\) cannot divide \(a^2 - a + 2\) unless \(p\) divides 2, which is a contradiction.
     - Therefore, \(p = 2\).

   - **Case 2: \(p = 2\).**
     - Since \(\gcd(a, a^2 - a + 2) = \gcd(a, 2) \leq 2\), we have \(\min\{\nu_p(a), \nu_p(a^2 - a + 2)\} = 1\).

     - **Subcase 1: \(\nu_p(a) = 1\).**
       - So \(a \equiv 2 \pmod{4}\), which implies \(4 \mid a^2 - a + 2\), hence \(k > 2\).
       - But \(k < 4d = 4\), so \(k = 3\).
       - We get \(a^2 + b + 1 = 2^3 = 8\), so \(a = 2\) and \(b = 3\).
       - However, in this case, \(a^2 + b + 1 = 8\) divides \(16 = (2 + 3 - 1)^2\), which contradicts the non-divisibility condition.

     - **Subcase 2: \(\nu_p(a^2 - a + 2) = 1\).**
       - Since \(p^k \mid a^2(a^2 - a + 2)\), we have \(2d + 1 \geq k\), hence \(d \geq \frac{k - 1}{2}\).
       - But we also have \(\frac{k}{2} > d\), hence we must have \(d = \frac{k - 1}{2}\) where \(k\) is odd.
       - Let \(k = 2n + 1\). So \(a = 2^n m\) where \(m\) is odd.
       - Since \(a^2 < 2^k\), we get \(2^{2n} m^2 < 2^{2n + 1}\), so \(m = 1\).
       - So we get \(a = 2^n\), \(b = 2^k - a^2 - 1 = 2^{2n + 1} - 2^{2n} - 1 = 2^{2n} - 1\).
       - It is easy to check that \((2^n, 2^{2n} - 1)\) satisfies the given conditions for all positive integers \(n > 1\) (for \(n = 1\) it's the same as case i).

Conclusion:
The pairs of positive integers \((a, b)\) that satisfy the given conditions are \((2^n, 2^{2n} - 1)\) for positive integer \(n > 1\).

The final answer is \( \boxed{ (2^n, 2^{2n} - 1) } \) for positive integer \(n > 1\).