Rewrite the equation as a quadratic in $x^2$
$$\begin{aligned} -1008 &= x^4 + y^4 + z^4 - 2x^2y^2 - 2y^2z^2 - 2z^2x^2 \\ &= x^4 - 2(y^2 + z^2)x^2 + (y^2 - z^2)^2. \end{aligned}$$
The discriminant is equal to $(y^2 + z^2)^2 - (y^2 - z^2)^2 = 4y^2z^2 = (2yz)^2$, hence the roots $x^2$ of the right hand side are $y^2 + z^2 \pm 2yz = (y \pm z)^2$. Therefore, the right hand side above factorises as
$$\begin{aligned} & (x^2 - (y+z)^2)(x^2 - (y-z)^2) \\ &= (x+(y+z))(x-(y+z))(x+(y-z))(x-(y-z)) \\ &= (x+y+z)(x-y-z)(x+y-z)(x-y+z). \end{aligned}$$
This implies $1008 = (x + y + z)(-x + y + z)(x + y - z)(x - y + z) = t \cdot u \cdot v \cdot w$, where we abbreviate $t = x + y + z$, $u = -x + y + z$, $v = x - y + z$ and $w = x + y - z$. We now have $t = u + v + w$ and $t \cdot u \cdot v \cdot w = 2^4 \cdot 3^2 \cdot 7$ and obtain $v + w = 2x$, $w + u = 2y$ and $u + v = 2z$, which implies that $u, v, w, t$ all have the same parity. As $1008$ is even, these four numbers must be even, i.e. there are integers $t', u', v', w'$ so that $t = 2t', u = 2u', v = 2v', w = 2w'$. From above these must satisfy

$$t' = u' + v' + w' \quad \text{and} \quad t' \cdot u' \cdot v' \cdot w' = 63 = 3^2 \cdot 7.$$
Therefore, $t'$ must be one of $1, 3, 7, 9, 21, 63$. But $t'$ cannot be $1$ or $3$, because in this case one of $u', v', w'$ would be divisible by $7$ which gives a sum larger than $t'$. On the other hand, $t'$ cannot be equal to $63$ or $21$ as in this case the sum of the remaining three factors would be at most $5$. So, we have to check only two cases: $t' = 7$ and $t' = 9$.
If $t' = 7$ we have $u' \cdot v' \cdot w' = 9$ and these three factors are either $1, 1, 9$ or $1, 3, 3$. Forming their sum we see that $(t', u', v', w') = (7, 1, 3, 3)$ or any permutation of the last three entries.
If $t' = 9$ we have $u' \cdot v' \cdot w' = 7$ and we obtain $(t', u', v', w') = (9, 1, 1, 7)$ or any permutation of the last three entries.
From above we obtain $x = v' + w'$, $y = w' + u'$ and $z = u' + v'$. This leads to the following six solution triples $(x, y, z)$:
(8, 8, 2)
(8, 2, 8)
(2, 8, 8)
(6, 4, 4)
(4, 6, 4)
(4, 4, 6)