We check that for $n = 3$
$$x^3 - 3x^2 + 2x + 6 = (x + 1)(x^2 - 4x + 6).$$
Suppose that for $n = 4$ we have
$$x^4 - 3x^3 + 2x^2 + 6 = (x^2 + a x + b)(x^2 + c x + d).$$
Then, comparing the coefficients we obtain
$$a + c = -3, \quad a c + b + d = 2, \quad b d = 6.$$
The first equation implies that $a$ and $c$ are of different parity and therefore, by the second equality, $b$ and $d$ are of the same parity. This contradicts the third equation. Hence we may assume that $n \ge 5$. Suppose that we have
$$W(x) = P(x) Q(x), \quad (1)$$
where
$$P(x) = a_k x^k + a_{k-1} x^{k-1} + \dots + a_1 x + a_0,$$
$$Q(x) = b_{n-k} x^{n-k} + b_{n-k-1} x^{n-k-1} + \dots + b_1 x + b_0,$$
and $a_k = b_{n-k} = \pm 1$. With no loss of generality we may assume that $k \le \lfloor \frac{1}{2} n \rfloor < n-2$ (because $n \ge 5$). Comparing the coefficients of both sides of (1) we obtain the following system of equations:
$$\begin{align*} a_0 b_0 &= 6, \\ a_0 b_1 + a_1 b_0 &= 0, \end{align*}$$
$$a_0 b_k + a_1 b_{k-1} + \dots + a_{k-1} b_1 + a_k b_0 = 0.$$
Now we can easily prove by induction that $a_0$ divides $a_1, a_2, \dots, a_k$. Indeed, having established this fact for $a_1, a_2, \dots, a_l$ we write
$$\begin{align*} 0 &= a_0(a_0 b_{l+1} + a_1 b_l + \dots + a_l b_1 + a_{l+1} b_0) = \\ &= a_0^2 b_{l+1} + a_0 a_1 b_l + \dots + a_0 a_l b_1 + 6 a_{l+1}, \end{align*}$$
and hence
$$6 a_{l+1} = -(a_0^2 b_{l+1} + a_0 a_1 b_l + \dots + a_0 a_l b_1).$$
We note that all the summands on the right hand side are divisible by $a_0^2$, therefore the left hand side also has this property. Hence $a_0 \mid a_{l+1}$.
But we have $a_k = \pm 1$; hence $a_0 = \pm 1$ and with no loss of generality we may take $a_0 = 1$ and then we have $b_0 = 6$.
Now if we repeat the above arguments for the coefficients of $Q$, then we see that $b_0 = 6$ divides $b_1, b_2, \dots, b_{n-3}$ (we set, if necessary, $b_l = 0$ if $l > n-k$). Then we obtain a contradiction ($b_{n-k} = \pm 1$), unless $n-k > n-3$. We consider two cases.
The case of $k=2$. Then we arrive at
$$a_0 b_{n-k} + a_1 b_{n-k-1} + \dots + a_{n-k-1} b_1 + a_{n-k} b_0 = 2,$$
a contradiction, because on the left hand side all the summands, except for the first one, are divisible by 6 and hence even, and the first summand is equal to $\pm 1$.
The case of $k=1$. Then the problem reduces to finding an integer root of the polynomial $W$; it is easy to check that if $n$ is even there are no such roots; if $n$ is odd, $W(-1) = 0$.
Therefore the answer to the problem is: $n$ is odd.