The short answer to your question is “ The TAS at 150 KIAS at FL300 is greater than the TAS at 150 KIAS at 5000 ft, so the power required at FL 300 is greater.
If that explanation is insufficient for your purposes a rather longer (but be no means the longest possible) answer is:
For an aircraft to fly at any given airspeed, work must be done to overcome the drag force.
That work is equal to the drag multiplied by the distance flown through the air.
Power is the rate of doing work.
So the power required to fly at any given airspeed is equal to the drag multiplied by the distance flown through the air divided by the time taken.
Distance flown divided by time taken = TAS
So Power Required = Drag x TAS
Drag = CD ½ Rho V squared where V is the TAS.
So drag is proportional to TAS squared.
So drag multiplied by TAS is proportional to TAS cubed (which is TAS squared x TAS )
So Power Required is proportional to TAS cubed.
The Indicated Airspeed (IAS) at any given altitude is determined by the TAS and the air density. As altitude increase the air density decreases, so more TAS is required to produce any given IAS.
At mean sea level in ISA conditions (ignoring instrument errors and pito system errors) the TAS is equal to IAS.
So at ISA MSL 150 KIAS = 150 Kts TAS.
But at 5000 ft in ISA conditions 150 KIAS = approximately 161 Kts TAS
And at FL300 in ISA conditions 150 KIAS = approximately 242 Kts TAS.
Looking at these last two figures, 242 kts is approximately 1.5 times greater than 161, So the power required at 150 KIAS at FL300 will be approximately 1.5 cubed = 3.375 times the power required at 150 KIAS at 5000 feet.