Let $a + a^2 + a^3 = n$, with integer $n$.
If $a$ is rational, then $a = \frac{p}{q}$, where $p, q$ are coprime integers. It follows from here that $\frac{p}{q} + \frac{p^2}{q^2} + \frac{p^3}{q^3} = n$, so $q^2p + qp^2 + p^3 = q^3n$, which implies $q \mid p^3$. This is possible only if $q = \pm 1$, which shows us that $a = \pm p$ is an integer number.

If $a^2$ is rational, then $a + a^3 = n - a^2$, so $a^2 + 2a^4 + a^6 = n^2 - 2a^2n + a^4$, which leads to $a^2(1 + 2n) + a^4 + a^6 = n^2$. An argument analogous to the previous one, applied to the last relation, shows us that $a^2$ is an integer and therefore $a$ is an integer.

If $a^3 = r$ is rational, then $a + a^2 = n - a^3$ is also rational; we get from here that the number $a^2 + 2a^3 + a^4$ is rational, so $a^2 + ra$ is also a rational number. Hence $a^2 + ra - (a^2 + a) = (r - 1)a$ is rational.
If $r \ne 1$, this would imply that $a$ is rational, so, due to the first part, it will follow from here that $a$ is an integer number. If $r = 1$, then $a^3 = 1$, which means $a = 1 \in \mathbb{Z}$.