Well, the second post there says that current doesn’t kill motors, so unsure how much you want to trust anything there.
He’s graphing some operating power (a steady state point), and not peak power of the motor at a given voltage.
Mechanical output power is going to be the speed times torque, and both are halved when voltage is halved… Halving voltage would put you at 25% peak power.
Neglecting things like heating, motor failure, etc… P1 = P2*(V1/V2)^2