I was curious to find out when, given $z_1,z_2\in \mathbb{C}$, I have that $z_1^{z_2}\in \mathbb{R}$ or $z_1^{z_2}\in \mathbb{I}$, where $\mathbb{I}$ is the set of imaginary numbers.
So I used the exponential notation and Euler's formula to obtain \begin{align} z_1^{z_2}&=r\exp(i\theta z_2)\\ &=r\exp(i\theta (a+ib))\\ &=r\exp(-\theta b)\exp(i\theta a)\\ &=r\exp(-\theta b)(\cos(\theta a) + i\sin (\theta a)). \end{align}
Then, by the sine and the cosine, I concluded that \begin{align} \left\{ \begin{aligned} z_1^{z_2} &\in \mathbb{R}&\quad\Leftrightarrow\quad&\frac{\arg (z_1)}{\pi}\mathfrak{Re}(z_2) \in \mathbb{Z}\\[0.7 em] z_1^{z_2} &\in \mathbb{I}&\quad\Leftrightarrow\quad&\frac{\arg (z_1)}{\pi}\mathfrak{Re}(z_2)+\frac{1}{2} \in \mathbb{Z}. \end{aligned} \right. \end{align}
First question: Is there any error in my conclusion?
Second question: If I take $z\in \mathbb{C}$, how can I visualize where are the values of $z$ such that $z^z\in \mathbb{R}$? What does it look like on the complex plane? I don't know how to plot something like that.
To answer your first question: No there is no error in your solution.
Now to answer: "What does $\{z\in \mathbb C\mid z^z\in \mathbb R\}$ look like?"
Well, it does not look very pretty. If we write $\arg(z)=\theta$, then $\Re(z)=r\cos \theta$. Rewriting what you found, this means that $$z^z\in \mathbb R\iff\frac{\theta}{\pi}r\cos\theta\in\mathbb Z\iff r=\left\vert\frac {k\pi}{\theta\cos\theta}\right\vert,$$ for some $k\in \mathbb Z$. (Note that we may assume $k>0$ without loss of generality.)
The best idea I have to get an idea of what this looks like is just plot it for different values of $k\in\mathbb Z$. Here are the plots in WolframAlpha for $k=1$ and $k=3$.
All of these plots depict points $z\in \mathbb C$, for wich $z^z\in \mathbb R$. If you draw all those plots (thus for all $k\in\mathbb Z_{\geq0}$) on the same sheet, then you will see $\left\{z\in\mathbb C\mid z^z\in\mathbb R\right\}$ appear.
As I said, this will not be "pretty".