Find all integers $n$ such that $n^4 + 8n + 11$ is a product of two or more consecutive integers.
To solve the problem of finding all integers \( n \) such that \( n^4 + 8n + 11 \) is a product of two or more consecutive integers, we can follow these steps:

1. **Check for the product of three or more consecutive integers:**
   - If \( n^4 + 8n + 11 \) is a product of three or more consecutive integers, it must be divisible by 3. 
   - We check the expression modulo 3:
     \[
     n^4 + 8n + 11 \pmod{3}
     \]
     - For \( n \equiv 0 \pmod{3} \):
       \[
       0^4 + 8 \cdot 0 + 11 \equiv 11 \equiv 2 \pmod{3}
       \]
     - For \( n \equiv 1 \pmod{3} \):
       \[
       1^4 + 8 \cdot 1 + 11 \equiv 1 + 8 + 11 \equiv 20 \equiv 2 \pmod{3}
       \]
     - For \( n \equiv 2 \pmod{3} \):
       \[
       2^4 + 8 \cdot 2 + 11 \equiv 16 + 16 + 11 \equiv 43 \equiv 1 \pmod{3}
       \]
   - Since \( n^4 + 8n + 11 \equiv 2 \pmod{3} \) or \( 1 \pmod{3} \), it is never divisible by 3. Therefore, \( n^4 + 8n + 11 \) cannot be a product of three or more consecutive integers.

2. **Check for the product of two consecutive integers:**
   - Let \( n^4 + 8n + 11 = a(a+1) \).
   - This implies:
     \[
     a^2 + a = n^4 + 8n + 11
     \]
   - Completing the square on the left-hand side:
     \[
     a^2 + a + \frac{1}{4} = n^4 + 8n + 11 + \frac{1}{4}
     \]
     \[
     \left(a + \frac{1}{2}\right)^2 = n^4 + 8n + 11 + \frac{1}{4}
     \]
     \[
     \left(a + \frac{1}{2}\right)^2 = n^4 + 8n + \frac{45}{4}
     \]
   - Let \( b = a + \frac{1}{2} \), then:
     \[
     b^2 = n^4 + 8n + \frac{45}{4}
     \]
     \[
     4b^2 = 4n^4 + 32n + 45
     \]
   - We need \( 4n^4 + 32n + 45 \) to be a perfect square. Let's check the range for \( n \geq 10 \):
     \[
     (2n^2)^2 < 4n^4 + 32n + 45 < (2n^2 + 1)^2
     \]
     \[
     4n^4 < 4n^4 + 32n + 45 < 4n^4 + 4n^2 + 1
     \]
   - For \( n \geq 10 \), the inequality holds, so we only need to check values for \( n \leq 9 \).

3. **Check values for \( n \leq 9 \):**
   - For \( n = 0 \):
     \[
     n^4 + 8n + 11 = 0^4 + 8 \cdot 0 + 11 = 11 \quad (\text{not a product of two consecutive integers})
     \]
   - For \( n = 1 \):
     \[
     n^4 + 8n + 11 = 1^4 + 8 \cdot 1 + 11 = 1 + 8 + 11 = 20 = 4 \cdot 5 \quad (\text{product of two consecutive integers})
     \]
   - For \( n = 2 \):
     \[
     n^4 + 8n + 11 = 2^4 + 8 \cdot 2 + 11 = 16 + 16 + 11 = 43 \quad (\text{not a product of two consecutive integers})
     \]
   - For \( n = 3 \):
     \[
     n^4 + 8n + 11 = 3^4 + 8 \cdot 3 + 11 = 81 + 24 + 11 = 116 \quad (\text{not a product of two consecutive integers})
     \]
   - For \( n = 4 \):
     \[
     n^4 + 8n + 11 = 4^4 + 8 \cdot 4 + 11 = 256 + 32 + 11 = 299 \quad (\text{not a product of two consecutive integers})
     \]
   - For \( n = 5 \):
     \[
     n^4 + 8n + 11 = 5^4 + 8 \cdot 5 + 11 = 625 + 40 + 11 = 676 = 26 \cdot 26 \quad (\text{not a product of two consecutive integers})
     \]
   - For \( n = 6 \):
     \[
     n^4 + 8n + 11 = 6^4 + 8 \cdot 6 + 11 = 1296 + 48 + 11 = 1355 \quad (\text{not a product of two consecutive integers})
     \]
   - For \( n = 7 \):
     \[
     n^4 + 8n + 11 = 7^4 + 8 \cdot 7 + 11 = 2401 + 56 + 11 = 2468 \quad (\text{not a product of two consecutive integers})
     \]
   - For \( n = 8 \):
     \[
     n^4 + 8n + 11 = 8^4 + 8 \cdot 8 + 11 = 4096 + 64 + 11 = 4171 \quad (\text{not a product of two consecutive integers})
     \]
   - For \( n = 9 \):
     \[
     n^4 + 8n + 11 = 9^4 + 8 \cdot 9 + 11 = 6561 + 72 + 11 = 6644 \quad (\text{not a product of two consecutive integers})
     \]

Conclusion:
The only integer \( n \) such that \( n^4 + 8n + 11 \) is a product of two or more consecutive integers is \( n = 1 \).

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