At full speed (which is only achieved when the motor is running free) torque is zero - so I think what you really mean is that you only get maximum torque at 100% PWM.
Torque is proportional to current, but you may wonder why current drops when the PWM ratio is reduced. The reason is that the average motor voltage drops, and maximum current draw is equal to voltage divided by resistance (at stall, when there is no back-emf so all the voltage appears across the resistance).
At 'full speed' (100% PWM) the motor voltage is highest, so if you want maximum torque you have to apply 100% PWM. But perhaps you want the motor to spin slower and still get high torque? In that case you need to increase the PWM ratio under heavy load.
One way to maintain full torque capability at low speed is develop a negative feedback loop based on rpm, which 'cancels out' the current limiting effect of the motor's internal resistance. Speed drops as loading increases because the higher current causes a greater voltage drop across the resistances in the circuit. Increasing the voltage (by raising the PWM ratio) compensates for this loss while it keeps the rpm constant.
Is applying a higher voltage OK?
Generally not. The motor is rated for a particular voltage based on speed and power loss. At higher speed there is more risk of bearing failure, excessive brush arcing and thrown windings (in a brushed motor) or thrown magnets (in a brushless motor), and higher magnetic losses. If you want to run at the same speed range as before then you must lower the PWM ratio - and you are back where you started except now you have higher switching losses.
Bottom line - if you are getting sufficient torque and rpm at 'full speed' (100% PWM) on your present setup then there is no need for higher voltage. If you aren't, and you would need to exceed the motor's voltage rating to get it, then you need a more powerful motor. If you already have more rpm than you need then consider using a gearbox, which trades speed for torque.