I think taking logs is an unnecessary indirection in the given proof. If n^m = m^n then raising both sides to the power of 1/(nm) gives us n^(1/n) = m^(1/m). So we are looking for two distinct positive integers at which the function x ↦ x^(1/x) takes the same value. The rest of the proof then goes as before.
Differentiating the above function yields (1/x^2)(1-log(x))x^(1/x), which is positive when log(x) < 1 and negative when log(x) > 1. So the function has a maximum at e and decreases on either side of it. Therefore one of our integers must be less than e, and the other greater than it. For the smaller integer there are only two possibilities, 1 and 2. Using 1 doesn't give a solution since the equation x^(1/x) = 1 only has the solution 1. So the only remaining possibility for the smaller number is 2, which does yield the solution 2^4 = 4^2. Since x^(1/x) is strictly decreasing when x > e, there can't be any other solutions with the same value.
Logs will appear no matter what, at some point in the line of reasoning. They are only held back until differentiating in this approach. I think the expression for the derivative of logn/n was much nicer to grapple with.
You often see this I think, in "pretty" proofs compared with the more direct approach. A clever early step or some bit of startling insight.