Substituting $a = b = c = 0$ yields $3f(0)^2 = 6f(0)^2$ which implies $f(0) = 0$. Now we can place $b = -a$, $c = 0$ to obtain $f(a)^2 + f(-a)^2 = 2f(a)f(-a)$, or, equivalently $(f(a) - f(-a))^2 = 0$ which implies $f(a) = f(-a)$.

Assume now that $f(a) = 0$ for some $a \in \mathbb{Z}$. Then for any $b$ we have $a+b+(-a-b) = 0$ hence $f(a)^2+f(b)^2+f(a+b)^2 = 2f(b)f(a+b)$, which is equivalent to $(f(b) - f(a+b))^2 = 0$, or $f(a+b) = f(b)$. Therefore if $f(a) = 0$ for some $a \neq 0$, then $f$ is a periodic function with period $a$.

Placing $b = a$ and $c = 2a$ in the original equation yields $f(2a) \cdot (f(2a) - 4f(a)) = 0$. Choosing $a = 1$ we get $f(2) = 0$ or $f(2) = 4f(1)$.

If $f(2) = 0$, then $f$ is periodic with period $2$ and we must have $f(n) = f(1)$ for all odd $n$. It is easy to verify that for each $c \in \mathbb{Z}$ the function
$$f(x) = \begin{cases} 0, & 2 \mid n, \\ c, & 2 \nmid n \end{cases}$$
satisfies the conditions of the problem.

Assume now that $f(2) = 4f(1)$ and that $f(1) \neq 0$. Assume that $f(i) = i^2 \cdot f(1)$ holds for all $i \in \{1, 2, \dots, n\}$ (it certainly does for $i \in \{0, 1, 2\}$). We place $a = 1$, $b = n$, $c = -n - 1$ in the original equation to obtain:
$$f(1)^2 + n^4 f(1)^2 + f(n+1)^2 = 2n^2 f(1)^2 + 2(n^2 + 1)f(n+1)f(1) \\ \Leftrightarrow (f(n+1) - (n+1)^2 f(1)) \cdot (f(n+1) - (n-1)^2 f(1)) = 0.$$
If $f(n+1) = (n-1)^2 f(1)$ then setting $a = n+1$, $b = 1-n$, and $c = -2$ in the original equation yields
$$2(n - 1)^4 f(1)^2 + 16f(1)^2 = 2 \cdot 4 \cdot 2(n - 1)^2 f(1)^2 + 2 \cdot (n - 1)^4 f(1)$$
which implies $(n-1)^2 = 1$ hence $n = 2$. Therefore $f(3) = f(1)$. Placing $a = 1$, $b = 3$, and $c = 4$ into the original equation implies that $f(4) = 0$ or $f(4) = 4f(1) = f(2)$. If $f(4) \neq 0$ we get
$$f(2)^2 + f(2)^2 + f(4)^2 = 2f(2)^2 + 4f(2)f(4)$$
hence $f(4) = 4f(2)$. We already have that $f(4) = f(2)$ and this implies that $f(2) = 0$, which is impossible according to our assumption.

Therefore $f(4) = 0$ and the function $f$ has period $4$. Then $f(4k) = 0$, $f(4k+1) = f(4k+3) = c$, and $f(4k+2) = 4c$. It is easy to verify that this function satisfies the requirements of the problem.

The remaining case is $f(n) = n^2 f(1)$ for all $n \in \mathbb{N}$, or $f(n) = cn^2$ for some $c \in \mathbb{Z}$. This function satisfies the given condition.

Thus the solutions are: $f(x) = cx^2$ for some $c \in \mathbb{Z}$; $f(x) = \begin{cases} 0, & 2 \mid n, \\ c, & 2 \nmid n \end{cases}$ for some $c \in \mathbb{Z}$; and $f(x) = \begin{cases} 0, & 4 \mid n, \\ c, & 2 \nmid n, \\ 4c, & n \equiv 2 \pmod{4} \end{cases}$ for some $c \in \mathbb{Z}$.