Suppose for the sake of contradiction that $n$ and $z$ are integers satisfying $f(n) = z^3$. Write $f(n) = 3(n^4 + n + 2) + n^4 - 7n^2$ and let $\tau \in \{0, 1, 2\}$ be the remainder of $n$ on division by 3.

Suppose 3 divides $z$. Then 3 divides $n^4 + 7n^2 = n^2(n^2 + 7)$. But $r^2 + 7 \in \{7, 8, 11\}$ is not divisible by 3. Hence 3 does not divide $n^2 + 7$ and so 3 must divide $n^2$. This implies that 3 divides $n$ and 9 divides $n^2(n^2 + 7)$. Since 9 divides $z^3$, it follows that 3 divides $n^4 + n + 2$. But, with $r$ as before, $r^4 + r^2 + 2$ is not divisible by 3, so 3 does not divide $n^4 + n + 2$, yielding a contradiction. Suppose that the remainder of $z$ on division by 3 is 1. Then, since 3 divides $f(n) - 1$, 3 must divide $n^4 + 7n^2 - 1$. Again, $r^4 + 7r^2 - 1$ is not divisible by 3, so this case is eliminated.
Finally, suppose $z$ leaves remainder 2 on division by 3. Then $z = 2 + 3h$, for some integer $h$, so $z^3 = 8 + 9k$, for some integer $k$. Now $n^4 + 7n^2 - 8 = (n^2 - 1)(n^2 + 8)$, and 3 must divide $n^2 - 1$ or $n^2 + 8$. But $n^2 + 8 - (n^2 - 1) = 9$, so both factors are divisible by 3, Hence, since $z^3 - 8$ is divisible by 9 also, 3 must divide $n^4 + n + 2$. But we have already shown this cannot happen. So we have reached a final contradiction, and the claimed result is established.

Note that the cubes modulo 9 are -1, 0, 1 and observe that $f(n) \equiv 4n^4 - 2n^2 + 3n - 3 = 2n^2(2n^2 - 1) + 3(n-1)$ (mod 9). The calculation modulo 9 in the table below shows that $f(n)$ is congruent to -4, -3 or 2 modulo 9, hence cannot be the cube of an integer.

| n | 0 | 1 | 2 | 3 | 4 | -4 | -3 | -2 | -1 |
|--------|----|----|----|----|----|----|----|----|----|
| n - 1 | -1 | 0 | 1 | 2 | 3 | 4 | -4 | -3 | -2 |
| 3(n-1) | -3 | 0 | 3 | -3 | 0 | 3 | -3 | 0 | 3 |
| 2n^2 | 0 | 2 | -1 | 0 | -4 | -4 | 0 | -1 | 2 |
| 2n^2(2n^2-1) | 0 | 2 | 2 | 0 | 2 | 2 | 0 | 2 | 2 |
| f(n) | -3 | 2 | -4 | -3 | 2 | -4 | -3 | 2 | -4 |

Therefore, $f(n)$ cannot be the cube of an integer.