DC Motor Efficiency at Fixed Torque
Some of your basic premise is correct : torque is proportional to current, and power DISSIPATED IN THE MOTOR ITSELF is a constant I^2*R, where R is the (constant) DC resistance of the motor, as measured across its terminals with the motor stationary.

Now let's run the motor at current I. The V required is not IR. (If it is, the motor is stalled so that the back EMF = 0.)

Instead, V = IR + back-EMF. 

Now, I * back-EMF is the electrical power delivered to the load as mechanical power, and I * IR is the power wasted in the motor as heat.

Let's increase V and increase speed keeping I constant. Now, input power has increased (IV) but the motor's resistance  hasn't changed : therefore IR is the same and I*IR is the same. But what HAS changed is the back-EMF - obviously, since it is proportional to speed (which has increased). 

So the power dissipated in the motor as heat is constant; but the electrical power delivered to the load (I * back-EMF) has increased, exactly as the mechanical output power (torque * speed) has.

No magic, and it all adds up correctly.

But what IS interesting is that the efficiency has increased because the wasted power is constant but the useful power has increased. So a general rule is that electrical efficiency is higher in a lightly loaded motor running fast, than a heavily loaded motor running slow and drawing high current. 

(There are limits to this : the less you load a motor and the faster you run it, the higher a proportion of power lost to friction in bearings and especially brushes. Bearings (ballraces) are easy : brushes are not, so brushless motors have a big advantage at high speed and high efficiency)