The maximum power transfer theorem tells you how to design a load to extract the maximum possible power from a given source. It does not tell you that the source will survive this treatment, and it does not tell you how to best design a source given a load. If you apply the maximum power transfer theorem to a typical solid-state RF transmitter, it will overheat (or reduce output to protect itself).
In a radio transmitter or amplifier, we design the output to operate efficiently, sending RF to the antenna and not heating itself up unnecessarily. A low impedance works fine for this purpose.
The maximum practical power is achieved at the point where the transmitter has reached its heat-dissipation limit (would overheat) or its circuits can no longer deliver more current to the output (would distort).
You may wonder: but don't we always make sure the antenna system is 50 Ω (or whatever impedance) everywhere? Aren't impedance mismatches really bad?
This is true — everywhere but at the transmitter/amplifier output! What we want to avoid is reflections that cause phase shifts. At the transmitter, there is not yet any length for the reflection to travel over, so there cannot be any phase shift. The output device is interacting instantaneously with (the matching network connected to) the 50 Ω line, and so the output can be designed as a lumped circuit driving a 50 Ω resistive load without transmission-line effects.
Another point is that this reasoning does not apply to receivers. In receivers, the incoming signal has already been traveling when it gets to the antenna and there is no active device driving it. An impedance mismatch at any point in the entire free space→antenna→feed line→receiver system will reflect some of the signal back out, making the received signal that much weaker.
Or, to say the same thing differently: the maximum power transfer theorem does apply to receivers, because your source (radio waves in free space) is unchangeable and so you match its impedance to extract as much signal power as possible.
(Unless, I suppose, you're operating in an environment of too much RF power, like some kind of RF-heating system or physics experiment. Then a deliberate mismatch could allow you to sample a calibrated fraction of the signal without overloading the receiver and without dissipating the RF in an attenuator.)
Of course, the receiver does not specifically need power in the sense of energy transferred over time (unless it is an RF-energy-harvesting device) — it needs a detectable signal. But that signal must be stronger than the noise inherent to the receiver circuit; and whether the receiver input is sensitive to voltage, current, or power, it will always see a better signal-to-noise ratio if less of the incoming signal is reflected.