Why does this driver not have a load resistor?
Why does this driver not have a load resistor?


Because it is not needed.

When a motor starts, for a short time there will be a starting current / inrush current / stall current.
When a motor is blocked or stalled, it will draw this stall current contineously.
If Q2 can handle this current, you don't need an additional load resistor.
If Q2 cannot handle this current, still do not use an additional load resistor because power is wasted in this resistor. Use a bigger transistor.

The TIP41 can handle up to 5A contineously, which is probably way more a small motor will draw.
So, leave out this 56&Omega; resistor.

To find out what stall current there will be and verify Q2 can handle it:
The motor's datasheet sometimes specifies this stall current at rated voltage or the motor winding resistance / terminal resistance. You can also measure the motor winding resistance with a multimeter. The stall current will be the applied motor voltage divided by this winding resistance.

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Regarding the circuit.
In general,if you just copy a paste a circuit and tweak some values, you cannot expect it to work properly.
So, just applying 5V and a circuit designed for 3.3V may damage it in 70% of the cases.
However, for this circuit, you can safely replace 3.3V by 5V. The 8051 has to sink 2mA instead of 1mA, but (I think) it is perfectly capable of that.
The circuit is working fine.
Don't short R3 as also indicated in one of the comments, it will damage both Q1 and Q2.