I can not find a good way to solve this rather simple-looking equation. cos⁡x+cos⁡2x=2 I...

You have already found all solutions.
The sum of those cosines can only be 2 if both x and ${\displaystyle \sqrt{2}x}$ are a multiple of ${\displaystyle 2\pi }$. Since ${\displaystyle \sqrt{2}}$ is not rational, there is no such multiple. In other words, the only solution is when:
${\displaystyle x=\sqrt{2}x=0\Rightarrow x=0}$

Use that
${\displaystyle \cos \left(x\right)+\cos \left(y\right)=2\cos (\frac{x}{2}-\frac{y}{2})\cos (\frac{x}{2}+\frac{y}{2})}$

From $-1\le \cos x\le 1,\mathrm{\forall }x\in \mathbf{R}$, we get $\cos x+\cos (\sqrt{2}x)\le 2$. The equality when $x=2k\pi$ : (1) and $\sqrt{2}x=2\lambda \pi$, where $k,\lambda \in \mathbf{Z}$. From (1) we get $\sqrt{2}(2k\pi )=2\lambda \pi \iff \frac{\sqrt{2}}{2}2k=2\lambda$, which is imposible when $k,\lambda$ integers and $k\ne 0(\sqrt{2}$ is irrational). Hence $\cos x+\cos (\sqrt{2}x)<2,\mathrm{\forall }x\in \mathbf{R}\mathbf{-}\{\mathbf{0}\}$. Hence the given equation has no real roots ecxept for the case $x=0$.