There are 4 solutions, namely
$$f(x) = x, \quad f(x) = 0, \quad f(x) = 2, \quad f(x) = \begin{cases} 1 & \text{if } x \text{ odd} \\ 2 & \text{if } x \text{ even} \end{cases}$$
Let (1) be the original equation. The first 3 solutions can be easily verified. The last solution can be verified by considering the cases when $x, y$ have the same parity, and when they don't. We proceed to prove these are the only solutions.

Put $x = y = 0$: $f(0)^2 = 2f(0) \Rightarrow f(0) = 0$ or $f(0) = 2$

Put $y = 0$: $f(x)^3 - 2f(x)^2f(0) + f(x)f(0)^2 + f(x)f(0) = f(x^3) + f(0)$ (2)

Put $y = -x$: $f(0)[(f(x) - f(-x))^2 + f(-x^2)] = f(x^3) + f(-x^3)$ (3)

Case 1. $f(0) = 0$. From (2) and (3) we get $f(x)^3 = f(x^3)$ and $f(x^3) = -f(-x^3)$. Therefore $f(-x) = \sqrt[3]{f(-x^3)} = \sqrt[3]{-f(x^3)} = -f(x)$, so $f$ is an odd function. Putting $x = 1$ in (2) yields $f(1)^3 = f(1)$, so $f(1) = 1, 0$ or $-1$. Putting $y = 1$ in (1), yields
$$f(x+1)((f(x)-f(1))^2 + f(x)) = f(x^3) + f(1) \quad (4)$$
Case 1.1. $f(1) = 0$. (4) becomes $f(x+1)(f(x)^2 + f(x)) = f(x)^3$. Thus $f(x+1) = 0$ implies $f(x) = 0$. Since $f(1) = 0$, $f(x) = 0$ for all $x \le 1$. Since $f$ is odd, $f(x) = 0$ for all $x \in \mathbb{Z}$ which is the second solution.

Case 1.2. $f(1) = 1$. (4) becomes $f(x+1)(f(x)^2 - f(x) + 1) = f(x)^3 + 1$. Since $f(x)^2 - f(x) + 1 > 0$, $f(x+1) = \frac{f(x)^3+1}{f(x)^2-f(x)+1} = f(x)+1$. From here we can easily show $f(x) = x$ for all $x \in \mathbb{Z}$ which is the first solution.

Case 1.3. $f(1) = -1$. (4) becomes $f(x+1)(f(x)^2 + 3f(x) + 1) = f(x)^3 - 1$. Since $f(x)^2 + 3f(x) + 1 > 0$, $f(x+1) = \frac{f(x)^3-1}{f(x)^2+3f(x)+1}$. Putting $x = 1$ and then $x = 2$ yield $f(2) = 2$ and $f(3) = \frac{7}{11}$ which is not an integer. Thus there is no solution.

Case 2. $f(0) = 2$. Substituting $x = 1$ in (2) yields $f(1)^3 - 4f(1)^2 + 5f(1) - 2 = 0$ which implies $(f(1) - 2)(f(1) - 1)^2 = 0$. Thus $f(1) = 1$ or $f(1) = 2$. Similarly, substituting $x = -1$ in (2) yields $f(-1) = 1$ or $2$. Thus $(f(1), f(-1))$ can be either $(1, 1)$, $(1, 2)$, $(2, 1)$, or $(2, 2)$. Putting $x = 1$ in (3), we get $2[(f(1) - f(-1))^2 + f(-1)] = f(1) + f(-1)$ implying $f(1) + f(-1)$ is even which rules out $(1, 2)$ and $(2, 1)$.

Putting $x = -1$ in (1), we get
$$f(x-1)((f(x)-f(-1))^2 + f(-x)) = f(x^3) + f(-1) \quad (5)$$
Case 2.1. $f(1) = f(-1) = 1$. We will inductively prove this leads to the function $f(\text{odd}) = 1$, $f(\text{even}) = 2$. It suffices to show the function is periodic with period 2. For $x \in [-1, 1]$ this is already true.

Suppose we have proven the periodicity for all $x \in [-n, n]$, where $n$ is a positive integer. This implies $f(n) = f(-n) > 0$. Since $f(1) = f(-1)$, combining (4) and (5) and substituting $x = n$, we get
$$f(n+1) = \frac{f(n)^3 + f(1)}{(f(n) - f(1))^2 + f(n)} = \frac{f(n)^3 + f(1)}{(f(n) - f(-1))^2 + f(-n)} = f(n-1)$$
Similarly, substituting $x = -n$, we get
$$f(-n-1) = \frac{f(-n)^3 + f(-1)}{(f(-n) - f(-1))^2 + f(n)} = \frac{f(-n)^3 + f(1)}{(f(-n) - f(1))^2 + f(-n)} = f(-n+1)$$
Thus the function is periodic with period 2 in the range $[-n - 1, n + 1]$ as well. The induction is complete and we have found our fourth solution.

Case 2.2. $f(1) = f(-1) = 2$. Similarly to Case 2.1, we can inductively prove the function is periodic with period 2, thus yielding the function $f(x) = 2$, the third solution.