$d = (a + 2^{\frac{1}{3}}b + 2^{\frac{2}{3}}c)^2 = a^2 + 4bc + 2^{\frac{1}{3}}(2ab + 2c^2) + 2^{\frac{2}{3}}(2ac + b^2)$. So by linear independence of $\{ 1, 2^{\frac{1}{3}}, 2^{\frac{2}{3}} \}$ over $\mathbb{Q}$, we must have $d = a^2 +4bc$, $2ab + 2c^2 = 0$, and $2ac + b^2 = 0$. The last two equations imply that $2a^2 = bc$ unless $bc = 0$, so $d = a^2 + 8a^2 = 9a^2 = (3a)^2$ or $d = a^2$is perfect square.

Let $\alpha = a + 2^{\frac{1}{3}}b + 2^{\frac{2}{3}}c$. Then $\alpha^2 = d$. Taking the norm $N: \mathbb{Q}(2^{\frac{1}{3}}) \rightarrow \mathbb{Q}$, we have $N(\alpha)^2 = d^3$. So $d^3$ is a perfect square, so $d$ is (note that $N(\alpha) \in \mathbb{Z}$ because it is integral over $\mathbb{Z}$ and is in $\mathbb{Q}$, and as $\mathbb{Z}$ is integrally closed the norm is an integer).