How close would a Sunlike star have to be in order to be dangerous to look at it through a spyglass? Alpha Centauri A which has an about 50% higher luminosity than the Sun is obviously undangerous to look at through a handheld telescope, at about 4.4 ly distance. How close would a yellow dwarf have to be in order to make it dangerous to look at it through an average spyglass? Or to feel some heat on the eye through the spyglass? This question can also be seen reversed as &quot;how far from the Sun would you have to be in order to look at the Sun through a spyglass without danger&quot;.
This is more a question of &quot;what is the brightest light I can look at&quot;, and subsequently this answer is more physiology, geometry than astronomy which only gives us the distances.
According to the wikipedia page on laser safety it is around 1mW/cm². A binocular - or any optics - again change the equation as so far that they accumulate all light falling onto its entry pupil into your eye.
Stars are sufficiently far away so that they are point sources (that's why they wobble around in the viewfinder due to atmospheric disturbance, unlike solar-system objects).
Energy emitted by a star is obtained by a simple black-body assumption $E=4\pi R^2_\star \sigma T^4$ and energy $W$ per on an area in distance $r$ thus is $W_\star = \frac{4\pi R^2_\star}{4\pi r^2}\sigma T^4 = \frac{L_\star}{4\pi r^2}$ (using the stellar luminosity $L_\star$ further on). Let's assume that your pupil has 0.5cm opening, thus the area there is $A_e = \pi 0.25^2 = 0.2cm^2 = 2\cdot 10^{-5}m^2$. Let's assume a binocular with 5cm aperture, thus the area there is $A_b=\pi\cdot 0.05^2m^2 = 8\cdot 10^{-3} m^2$, thus the ratio $A_e / A_b = 2.5\cdot 10^{-3}$. Thus the light at the entry of the binocular needs to be $W_{max} = 1mW/cm^2 \cdot 2.5\cdot 10^{-3} = 2.5\mu W/cm^2 = 2.5\cdot 10^{-2} W/m^2$ at most to not become dangerous. So, what distance does that imply? Let's equate:
$W_{max} = W_\star $ and solve for $r$: $r = \sqrt{\frac{L_\star}{4\pi W_{max}}}$. So for a sun-like star ($L_\star = 4\cdot 10^{26} W$) we get a minimum distance to safely look at its disk of roughly r=240 AU or 0.004 light years.
You can modify that equation for other stars with higher or lower energy output. Alpha centauri is not much brighter than our sun. But for a star like Vega ($L=40L_{sun}$), the minimum safe distance would already be more than 6 times as distant - yet still very near for any distances when considering distances to other stars. Thus any star becoming dangerous to look at with the naked eye or a binocular would pose a threat to our solar system.
And even if you consider not binoculars, but telescopes, say 1m diameter, thus 400x the light gathering power compared to the considered binocular, we would start to get distances of the order of a light year... even then all stars are further away.
(this still sounds crazy... is there an error in the estimates? But likely this errs on the side of caution. While you should not look at the sun even with the unaided eye, you're not immediately blind - and it's intensity is $1400W/m^2$ compared to the assumed safe power of $0.025W/m^2$, thus there's a difference of $10^5$)