Let $z\in \mathbb C$ such that $Re (z) >0 $ and $Im(z)\ne 0$. Then must there exist integer $n>1$ such that $Re (z^n) <0$ ?
Equivalently, given $\theta \in [0, 2\pi)$ such that $\cos \theta >0$ and $\sin \theta \ne 0$ , must there exist $n>1$ integer such that $\cos (n\theta) <0$ ?
Yes, it's true.
Without loss of generality, we focus on $z$ with $|z| = 1$, and write $z = \cos(\theta) + i \sin(\theta)$, with $\theta \in (0, \frac\pi2) \cup ( \frac{3\pi}{2}, 2\pi)$ (since you're excluding real $z$, and only allowing $z$ with positive real part). Recall that by De Moivre, we have $z^k = \cos(k \theta) + i \sin(k \theta)$.
Note that the complex conjugate $\bar{z}$ has the same real part as $z$, and the same goes for $\bar{z}^n = \overline{z^n}$. For this reason, $z$ has some power with negative real part if and only if $\bar{z}$ does. Thus, we may further suppose that $\theta \in (0, \frac\pi2)$.
The basic idea is that $\theta$ must be between two angles of the form $\frac\pi n$ and $\frac{\pi}{n+1}$ for large enough $n$, and these are (relatively) easily shown to have a common multiple in the 2nd quadrant, hence some multiple of $\theta$ is there too.
Because $\theta \in (0, \frac\pi2)$, there exists some $n$ so that $\frac{\pi}{n+1} \le \theta < \frac{\pi}{n}$ (this is equivalent to an $n$ such that $n < \frac{\pi}{\theta} \le n + 1$), where $n > 2$.
Consequently, $\frac{n}{n+1}\pi \le n \theta < \pi$, where $$\frac{n}{n+1}\pi = (1 - \frac{1}{n+1})\pi > (1 - \frac{1}{n})\pi > \frac\pi2$$ since $n > 2$.
Thus, $\frac{\pi}{2} < n\theta < \pi$, hence $\cos(n\theta)$, the real part of $z^n$, is negative.