I need to find $x \in \mathbb{R}$ such that:
$$\sum^{\infty}_{n=1} \frac{\cos(nx)}{e^{nx}}$$
is pointwise convergent and characterise sets on which the series is uniformly convergent.
Now, I think that the series is pointwise convergent for every $x > 0$. Because in that case:
$$\sum^{\infty}_{n=1} \frac{\cos(nx)}{e^{nx}} \leq \sum^{\infty}_{n=1} \frac{1}{e^{nx}} \leq \sum^{\infty}_{n=1} \frac{1}{n^2} \rightarrow \text{converges}$$
If $x = 0$ then we get: $$\sum^{\infty}_{n=1} \frac{\cos(nx)}{e^{nx}} = \sum^{\infty}_{n=1} 1 = \infty$$
if $x < 0$:
$$\sum^{\infty}_{n=1} \frac{\cos(nx)}{e^{-nx}} = \sum^{\infty}_{n=1} e^{nx}\cos(nx) = \infty$$
I don't know how to find sets on which the series is uniformly convergent.
Your comparison test above for pointwise convergence is not valid since the cosine term changes sign. But you can easily fix this as an absolute convergence comparison test, since $\left|\frac{\cos nx}{e^{nx}} \right| \leqslant e^{-nx}$ and for $x > 0$ the series $\sum e^{-nx}$ is a convergent geometric series.
If $x \in [a,\infty)$ where $a > 0$, then $\displaystyle\left|\frac{\cos n x}{e^{nx}} \right| \leqslant e^{-na}$ and the series converges uniformly by the Weierstrass M-test on $[a,\infty)$.
Obviously, the series diverges for any $x \leqslant 0$ as you showed.
Finally we can show that the series fails to converge uniformly on $(0,\infty)$. Note that for any $n \in \mathbb{N}$ and $x_n = \pi/(6n)\in (0,\infty)$, we have $1/2 < \cos kx_n \leqslant \sqrt{3}/2$ for $n < k \leqslant 2n$, and
$$\left|\sum_{k=n+1}^{2n}\frac{\cos kx_n}{e^{kx_n}}\right| = \sum_{k=n+1}^{2n}\frac{\cos kx_n}{e^{kx_n}} > n \cdot \frac{1/2}{e^{2nx_n}} = \frac{n}{2e^{\pi/3}} \underset{n \to \infty}\longrightarrow\infty,$$
and the series fails to converge uniformly on $(0,\infty)$ by violation of the uniform Cauchy criterion.