We will prove that $n^{4}+8 n+11$ is never a multiple of 3. This is clear if $n$ is a multiple of 3. If
$n$ is not a multiple of 3, then $n^{4}+8 n+11=\left(n^{4}-1\right)+12+8 n=(n-1)(n+1)\left(n^{2}+1\right)+12+8 n$, where $8 n$ is the only term not divisible by 3. Thus $n^{4}+8 n+11$ is never the product of three or more integers.

It remains to discuss the case when $n^{4}+8 n+11=y(y+1)$ for some integer $y$. We write this as $4\left(n^{4}+8 n+11\right)=4 y(y+1)$ or $4 n^{4}+32 n+45=(2 y+1)^{2}$. A check shows that among $n= \pm 1$ and $n=0$ only $n=1$ satisfies the requirement, as $1^{4}+8 \cdot 1+11=20=4 \cdot 5$. Now let $|n| \geq 2$. The identities $4 n^{2}+32 n+45=\left(2 n^{2}-2\right)^{2}+8(n+2)^{2}+9$ and $4 n^{4}+32 n+45=$ $\left(2 n^{2}+8\right)^{2}-32 n(n-1)-19$ indicate that for $|n| \geq 2,2 n^{2}-2<2 y+1<2 n^{2}+8$. But $2 y+1$ is odd, so it can equal $2 n^{2} \pm 1 ; 2 n^{2}+3 ; 2 n^{2}+5$ or $2 n^{2}+7$. We investigate them one by one.

If $4 n^{4}+32 n+45=\left(2 n^{2}-1\right)^{2} \Rightarrow n^{2}+8 n+11=0 \Rightarrow(n+4)^{2}=5$, which is impossible, as 5 is not a perfect square.

If $4 n^{4}+32 n+45=\left(2 n^{2}+1\right)^{2} \Rightarrow n^{2}-8 n-11=0 \Rightarrow(n-4)^{2}=27$ which also fails.

Also $4 n^{4}+32 n+45=\left(2 n^{2}+3\right)^{2} \Rightarrow 3 n^{2}-8 n-9=0 \Rightarrow 9 n^{2}-24 n-27=0 \Rightarrow(3 n-4)^{2}=43$ fails.

If $4 n^{4}+32 n+45=\left(2 n^{2}+5\right)^{2} \Rightarrow 5 n^{2}-8 n=5 \Rightarrow 25 n^{2}-40 n=25 \Rightarrow(5 n-4)^{2}=41$ which also fails.

Finally, if $4 n^{4}+32 n+45=\left(2 n^{2}+7\right)^{2}$, then $28 n^{2}-32 n+4=0 \Rightarrow 4(n-1)(7 n-1)=0$, whence $n=1$ that we already found. Thus the only solution is $n=1$.

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 $\boxed{ n = 1 }$.