Am I doing this right?
$e^{2z}+e^z +1=0$. Let $x=e^z$ so the original equation translates to \begin{align}x^2+x+1=0\end{align} Using the quadratic formula for real numbers, we get $\frac{-1}{2}+\frac{i\sqrt{3}}{2}, \frac{-1}{2}+\frac{i\sqrt{3}}{2}$. Equating $e^z$ with both values, let $z=r(\cos\theta +i\sin\theta)$ and we have \begin{align} e^z=e^{x+iy}=e^x(\cos y+i\sin y)&=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=\frac{-1}{2}+\frac{i\sqrt{3}}{2}\\ e^x=1 \quad \text{and}\quad y&=\frac{2\pi}{3}+2\pi k\quad k \in \Bbb{Z}\\ x&=0 \quad \text{and} \quad y=\frac{2\pi}{3}\\ \text{Also,} \quad e^x(\cos y+\sin y)&=\cos(\frac{4\pi}{3})+i\sin\frac{4\pi}{3})=\frac{-1}{2}+\frac{i\sqrt{3}}{2}\\ e^x=1 \quad \text{and}\quad y&=\frac{4\pi}{3}+2\pi k\quad k \in \Bbb{Z}\\ x&=0 \quad \text{and} \quad y=\frac{4\pi}{3}\\ \end{align} Therefore, the solutions are this values, again, $\frac{-1}{2}+\frac{i\sqrt{3}}{2}, \frac{-1}{2}+\frac{i\sqrt{3}}{2}$.
Beware the notations ($x$) clash. $\color{red}{\text{Suppose there exists }} z \in \mathbb{C} \text{ such that } e^{2z}+e^z+1=0.$ Let $u=e^z$. $$u^2+u+1=0$$Then, $e^z=u$, with $u\in\{e^{i\frac{2\pi}{3}},e^{i\frac{4\pi}{3}}\}$. Then, $\color{red}{\text{necessarily, }}z=i\left(\frac{2\pi}{3}+2n\pi\right) \text{ or } i\left(\frac{4\pi}{3}+2n\pi \right)$, with $n\in \mathbb{Z}.$
Conversely, $\color{red}{\text{these}}$ solutions are suitable:$\{z\in\mathbb{C}:e^{2z}+e^z+1=0\}=\{i\left(\frac{2\pi}{3}+2n\pi\right),i\left(\frac{4\pi}{3}+2n\pi \right):n\in \mathbb{Z}\}$