I suppose that Mathematica makes the following simplification In[103]:= Simplify[(8^x - 2^x)/(6^x - 3^x)]

Out[103]= (2/3)^x (1 + 2^x)

and then does not check back the original equation. This is a well-known issue:

In[104]:= Solve[(x^2 - 1)/(x - 1) == 2, x]

Out[104]= {{x -> 1}}

However, just plot the function

Plot[(8^x - 2^x)/(6^x - 3^x) - 2, {x, -1, 2}]

and you will see that Mathematica's answer is not so incorrect in some sense.

FunctionDomain checks back on the original form of the function. Try this:

Solve[(8^x - 2^x)/(6^x - 3^x) == 2 && 
  FunctionDomain[(8^x - 2^x)/(6^x - 3^x), x], x, Reals]

I am not familiar with the workings of W|A in this case, but Solve will handle removable singularities by removing them.

I didn't claim solutions at singularities would be removed. I stated that the singularities themselves would be removed, that is, not be regarded as singularities (that's what it means to remove a removable singularity). Stated in terms of the most recent example, in the field of rational functions (x^2-1)/(x-1) is equivalent to (x+1).

This is not regarded as broken by the way. Wolfram|Alpha is more in tune with what is taught in high schools, and will attempt to remove singular solutions. So that particular example will have an empty solution set. Different philosophy.

You are correct, it is only removing the algebraic ones, that is, those that arise from rational functions. It is not removing more general ones from meromorphic functions.