Why does QAM use two cycles instead of one per byte?
It doesn't matter how long you stay on one constellation point. The constellation diagram has NO ties to time whatsoever. The two base vectors $e_x, e_y$ (x-axis and y-axis) represent your real and imaginary part or "basic functions" your signal is made of. For example, if your signal consists of two functions (in this case, QAM, it consists of a real and an imaginary part including the amplitude)

$$
e_x = A\cos(2\pi f_{carrier}t) \\
e_y = A\sin(2\pi f_{carrier}t) \\
$$

If you stay on one constellation point for an hour, that's fine. The constellation diagram only shows you which "constellation" of amplitude and phase you should choose to transmit a specific "constellation" of bits (001 010 110, etc ..).

The question you ask is about the symbol rate. If the symbol rate is half your carrier frequency, then you will always have two cycles before the next symbol will be sent. If your symbol frequency is a quarter of your carrier frequency, you will have four cycles to the next symbol.

I think there are always at least one and a half cycle shown to make it easier on the eye and for the brain to understand because we can follow the sine easily. This wouldn't be the case if you'd change the phase and amplitude every cycle. That being said, there's also the problem of having enough time for the demodulator to detect the phase shifts. This is why the symbol frequency is often a lot lower than the carrier frequency.