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]