How high can an airplane be spotted with eyesight?
Note: There is some confusion in the comments about this answer. What I am answering here is the question:


  Suppose a plane has a wingspan of 15 m then at what distance it will appear as a faint dot? What is the correct way to calculate this?


This does not imply that you cannot see the airplane at a greater distance any more. It just means you cannot resolve it any more, i.e. it will appear as a faint dot.



This can be estimated with the Rayleigh criterion. The angular resolution of the human eye is given by

$$ \theta = 1.22 \frac{\lambda}{D} $$

where $ \lambda $ is the wavelength of the light and $ D $ is the pupil diameter. Using typical values of $ \lambda = 550 \, \mathrm{nm} $ (highest sensitivity of the human eye) and $ D = 3 \, \mathrm{mm} $ during the day, we get a resolution of

$$ \theta \approx 2.24 \times 10^{-4} \, \mathrm{rad} \approx 0.013^\circ $$

If the aircraft is $ w = 15 \, \mathrm{m} $ wide, the distance at which this corresponds to this angle is given by

$$ d = \frac{w}{\tan \theta} \approx \frac{w}{\theta} \approx 67.1 \, \mathrm{km} $$

Most planes only fly up to FL410, which is about 12.5 km, so you could always distinguish them from a faint point with the naked eye. As Zeiss Ikon pointed out in the comments, the actual resolution of your eyes may be worse because your eyesight is probably not perfect. This is however an upper limit on the resolution.