Find all integers $x$ such that $2x^2+x-6$ is a positive integral power of a prime positive integer.
1. We start with the given quadratic expression:
   \[
   2x^2 + x - 6
   \]
   We need to find all integers \( x \) such that this expression is a positive integral power of a prime positive integer.

2. First, factor the quadratic expression:
   \[
   2x^2 + x - 6 = (2x - 3)(x + 2)
   \]
   This factorization can be verified by expanding:
   \[
   (2x - 3)(x + 2) = 2x^2 + 4x - 3x - 6 = 2x^2 + x - 6
   \]

3. We need \( (2x - 3)(x + 2) \) to be a positive integral power of a prime positive integer. Let \( p^k \) be this power, where \( p \) is a prime and \( k \) is a positive integer.

4. Consider the possible values of \( 2x - 3 \) and \( x + 2 \):
   - Both factors must be integers.
   - Both factors must be positive or both must be negative (since their product is positive).

5. We analyze the cases where \( 2x - 3 \) and \( x + 2 \) are equal to each other, equal to 1, or equal to -1.

6. Case 1: \( 2x - 3 = x + 2 \)
   \[
   2x - 3 = x + 2 \implies 2x - x = 2 + 3 \implies x = 5
   \]
   Check:
   \[
   2(5)^2 + 5 - 6 = 50 + 5 - 6 = 49 = 7^2
   \]
   So, \( x = 5 \) is a solution.

7. Case 2: \( 2x - 3 = 1 \) and \( x + 2 = 1 \)
   \[
   2x - 3 = 1 \implies 2x = 4 \implies x = 2
   \]
   Check:
   \[
   2(2)^2 + 2 - 6 = 8 + 2 - 6 = 4 = 2^2
   \]
   So, \( x = 2 \) is a solution.

8. Case 3: \( 2x - 3 = -1 \) and \( x + 2 = -1 \)
   \[
   2x - 3 = -1 \implies 2x = 2 \implies x = 1
   \]
   Check:
   \[
   2(1)^2 + 1 - 6 = 2 + 1 - 6 = -3 \quad (\text{not a positive power of a prime})
   \]
   So, \( x = 1 \) is not a solution.

9. Case 4: \( 2x - 3 = -1 \) and \( x + 2 = -1 \)
   \[
   x + 2 = -1 \implies x = -3
   \]
   Check:
   \[
   2(-3)^2 + (-3) - 6 = 18 - 3 - 6 = 9 = 3^2
   \]
   So, \( x = -3 \) is a solution.

10. We have found the solutions \( x = 5, 2, -3 \).

The final answer is \(\boxed{-3, 2, 5}\)