Are there $x,y,z \in \mathbb Q \left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3}\right)\right)$ such that $x^2+y^2+z^2=-1$?

Let $\omega = \exp\left(\frac{2\pi i}{3}\right)$. As you noted $\Bbb{Q}[\omega\sqrt[3]{2}]$ is a root or stem field for $x^3-2$, so that we have an isomorphism $\sigma: \Bbb{Q}[\omega\sqrt[3]{2}]\mapsto \Bbb{Q}[\sqrt[3]{2}]$. Suppose that there were $x,y$ and $z$ in $\Bbb{Q}[\omega\sqrt[3]{2}]$ such that $x^2+y^2+z^2=-1$, then we would have $\sigma(x)^2+\sigma(y)^2+\sigma(z)^2=-1$. But this last equation is obviously impossible, since $\sigma(x), \sigma(y)$ and $\sigma(z)$ are real numbers.

This answer is based on @sharding4's and @JyrkiLahtonen's comments. I hope I formalized everything appropriately, give or take a few hand-waving explanations.

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Let $g(x,y,z) = x^2+y^2+z^2+1$. $$E = \left\{e_1 = \sqrt[3]{2} \exp\left( \frac{2\pi i}{3}\right),\ e_2=e_1^2=\sqrt[3]{4} \exp\left( \frac{4\pi i}{3}\right), \ e_3=e_1^3 = 2 \right\}$$ is a basis of $\mathbb{Q}\left(\sqrt[3]{2} \exp\left( \frac{2\pi i}{3}\right)\right)$. Also, $$F=\left\{f_1 = \sqrt[3]{2},\ f_2=f_1^2=\sqrt[3]{4}, \ f_3=f_1^3 = 2 \right\}$$ is a basis of $\mathbb{Q}\left(\sqrt[3]{2}\right)$.

Let $\varphi \left(\sum_{i=1}^3 q_i\cdot e_i \right) = \sum_{i=1}^3 q_i\cdot f_i$. It's easy to see that $$\varphi:\mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3}\right)\right)\to\mathbb{Q}\left(\sqrt[3]{2}\right)$$ is an isomorphism of fields. Hence, there are $x,y,z\in \mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3}\right)\right)$ such that $g(x,y,z)=0$ if and only if $$\varphi \left(g\left(x,y,z\right)\right)=g\left(\varphi \left(x\right),\varphi \left(y\right),\varphi \left(z\right)\right)=0.$$ But $g(a,b,c)\ge 1$ for all $a,b,c\in\mathbb{Q}\left(\sqrt[3]{2}\right)\subset\mathbb{R}$. Therefore, there are no $x,y,z \in \mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3}\right)\right)$ such that $x^2+y^2+z^2=-1$.