We have $x^{12} + 1 = (x^4 + 1)(x^8 - x^4 + 1)$. It is easy to check that these two factors are irreducible in $\mathbb{Z}[x]$ (or equivalently in $\mathbb{Q}[x]$). On the other hand, $P^3 + Q^3 = (P+Q)(P^2-PQ+Q^2)$. Therefore, our goal is to find $P$ and $Q$ such that $(P+Q)(P^2-PQ+Q^2) = (x^4+1)(x^8-x^4+1)$. Irreducibility of $x^4+1$ and $x^8-x^4+1$ imply
* $x^4 + 1$ divides $P + Q$ or $P^2 - PQ + Q^2$.
* $x^8 - x^4 + 1$ divides $P + Q$ or $P^2 - PQ + Q^2$.
So we have four cases
* $P + Q = 1$ and $P^2 - PQ + Q^2 = x^{12} + 1$.
$$1 - 3PQ = (P+Q)^2 - 3PQ = P^2 - PQ + Q^2 = 1 + x^{12} \Rightarrow 3P(1-P) = -x^{12}$$
So zero is the unique root of both $P$ and $1-P$ which is impossible.
* $P + Q = x^{12} + 1$ and $P^2 - PQ + Q^2 = 1$. Same as the previous part, we get $3PQ = x^{12}(x^{12} + 2)$. So $P(0) = 0$ or $Q(0) = 0$, but at most one of them can be zero because $P + Q = x^{12} + 1$. Because of symmetry we assume that $P$ is divisible by $x^{12}$. Now since $3PQ = x^{12}(x^{12} + 2)$, we conclude $\deg(Q) \le 12$. If $\deg(Q) < 12$, then degree of $P$ must be greater than 12 and so we cannot have $P + Q = x^{12} + 1$. Therefore, $\deg(P) = \deg(Q) = 12$. Hence, there must be rational numbers $a$ and $b$ such that $P(x) = a x^{12}$ and $Q(x) = b(x^{12} + 2)$. So $P + Q = (a + b)x^{12} + 2b = x^{12} + 1$. Thus, $a = b = \frac{1}{2}$. This is not possible because $3PQ = 3abx^{12}(x^{12} + 2) = \frac{3}{4}x^{12}(x^{12} + 2) \neq x^{12}(x^{12} + 2)$. Hence, there is no solution in this case.
* $P + Q = x^8 - x^4 + 1$ and $P^2 - PQ + Q^2 = x^4 + 1$. In this case we have $3PQ = x^4(x^{12} - 2x^8 + 3x^4 - 3)$ and by arguments similar to the previous part, we deduce that $\deg(P) = \deg(Q) = 8$. Let $p$ and $q$ be the leading coefficients of $P$ and $Q$, respectively. We have
$$\begin{aligned} P + Q &= x^8 - x^4 + 1 & \Rightarrow p + q &= 1 \\ 3PQ &= x^4(x^{12} - 2x^8 + 3x^4 - 3) & \Rightarrow pq &= \frac{1}{3} \end{aligned}$$
But $4pq = \frac{4}{3} > 1 = (p+q)^2$ and so $p$ and $q$ are not real numbers. Thus, we do not have any solutions in this case.
* $P+Q = x^4+1$ and $P^2-PQ+Q^2 = x^8-x^4+1$. We get $PQ = x^4$. Note that again we cannot have $P(0) = Q(0) = 0$, because $P(0) + Q(0) = 1$. So we can assume that for example $P$ is divisible by $x^4$ and so $Q$ is a constant polynomial. Suppose that $P(x) = p x^4$ and $Q(x) = q$ ($p, q \in \mathbb{Q}$). Now since $P + Q = p x^4 + q = x^4 + 1$ we get $p = q = 1$. This leads to the solution $P(x) = x^4$ and $Q(x) = 1$. $P(x) = 1$ and $Q(x) = x^4$ is another solution.