Find all positive integers $n$ and prime numbers $p$ such that $$17^n \cdot 2^{n^2} - p =(2^{n^2+3}+2^{n^2}-1) \cdot n^2.$$

[i]Proposed by Nikola Velov[/i]
1. **Initial Claim and Setup**:
   We claim that the only pair \((p, n)\) that satisfies the equation is \((17, 1)\).

2. **Case \(p = 2\)**:
   \[
   17^n \cdot 2^{n^2} - 2 = (2^{n^2+3} + 2^{n^2} - 1) \cdot n^2
   \]
   - Since \(p = 2\), we need to check if \(n\) can be even or odd.
   - If \(n\) is even, then \(4 \mid \text{RHS}\) because \(2^{n^2+3}\) and \(2^{n^2}\) are both divisible by \(4\).
   - However, for the LHS, \(v_2(17^n \cdot 2^{n^2} - 2) = 2\), which is a contradiction because the LHS is not divisible by \(4\).

3. **Case \(p > 2\)**:
   - Here, \(n\) must be odd because if \(n\) were even, the LHS and RHS would not match in terms of divisibility by \(4\).
   - Since \(n\) is odd, \(n^2 \equiv 1 \pmod{8}\).

4. **Modulo 17 Analysis**:
   - We inspect both sides modulo \(17\):
     \[
     2^{n^2+3} \equiv -1 \pmod{17} \quad \text{and} \quad 2^{n^2} \equiv 2 \pmod{17}
     \]
     - Therefore, the RHS modulo \(17\) is:
       \[
       2^{n^2+3} + 2^{n^2} - 1 \equiv -1 + 2 - 1 \equiv 0 \pmod{17}
       \]
     - This implies \(17 \mid \text{RHS}\), so \(p = 17\).

5. **Simplifying for \(p = 17\)**:
   - Substituting \(p = 17\) into the equation:
     \[
     17 \cdot (17^{n-1} \cdot 2^{n^2} - 1) = (2^{n^2+3} + 2^{n^2} - 1) \cdot n^2
     \]

6. **Analyzing for \(n \geq 3\)**:
   - For \(n \geq 3\), we compare the magnitudes of both sides:
     \[
     17^{n-1} > 9n^2
     \]
     - This implies:
       \[
       17^{n-1} \cdot 2^{n^2} - 1 > 9n^2 \cdot 2^{n^2} - 1
       \]
     - Therefore:
       \[
       \text{LHS} > 17 \cdot (9n^2 \cdot 2^{n^2} - 1)
       \]
     - On the other hand:
       \[
       \text{RHS} = n^2 \cdot (9 \cdot 2^{n^2} - 1) = 9n^2 \cdot 2^{n^2} - n^2 < 17 \cdot (9n^2 \cdot 2^{n^2} - 1)
       \]
     - This is a contradiction for \(n \geq 3\).

7. **Conclusion for \(n = 1\)**:
   - Since \(n\) must be odd and \(n \geq 3\) leads to a contradiction, we check \(n = 1\):
     \[
     17^1 \cdot 2^{1^2} - 17 = (2^{1^2+3} + 2^{1^2} - 1) \cdot 1^2
     \]
     - Simplifying:
       \[
       34 - 17 = (2^4 + 2 - 1) \cdot 1
       \]
       \[
       17 = 17
       \]
     - This holds true.

The final answer is \( \boxed{ (p, n) = (17, 1) } \).