Under real-world optics, it's impossible. The upper limit of visibility through an Earth-like atmosphere in best possible conditions is about 300 km. Wiki gives a technical explanation of how this is calculated, which I'll summarise:
As light passes through the air, it interacts with stuff and gets absorbed or scattered (think "deflected"). That "stuff" can be particulates (smoke, fog etc. in the air) but even in perfectly clean air, light will interact with the nitrogen and oxygen molecules of the air itself.
The result of that air interaction (Rayleigh scattering) is that about 1.32% of light is scattered for every kilometer between the object and its observer.
This scattering reduces contrast, and past about 300 km so much contrast is lost that it's impossible to distinguish between objects and their surroundings.
What if the mountain is so high that it sticks up above the atmosphere? Most of the Earth's atmosphere is in the bottom 10 km or so. Taking that as an approximate cutoff, a little bit of trigonometry indicates that if our line of sight to the top of the mountain passes through no more than 300km of atmosphere, then the top of the mountain needs to be about (1290/300)*10 = 43 km high, or about five times as high as Everest.
IRL, geology suggests that Everest is about as high as a mountain on Earth can get before erosion and its own mass pulls it down.
The above assumes that our observer is at sea level. What if we're standing on top of one Everest-like mountain, looking at another?
Air density at the peak of Everest is about 0.30 x that at sea level. (Based on atmospheric pressure; should probably be a little higher due to colder temperatures, but let's ignore that.) If we assume a constant density of 0.30 x surface for the line of sight between the two peaks, then we might be able to extend as far as 300/0.30 = 1000 km, but this still falls a little short of the 1290 km required.
If you're willing to ignore scattering (perhaps the same magic that reshaped the world also changed the way optics works?), then L. Dutch's answer based on resolution of 1 minute of arc is good.