Transistor on shutdown pin
Your illustration is a wiring diagram and, unfortunately, doesn't show us the scheme of the circuit. For example, which pin is the base on the transistors?
 Why are there two? We need something more like this:







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Simple level converter.


  ... and consumed about 1ma. Which I think is exactly what you would expect through the 10kΩ pullup.



The 10k pull-up will pass \$ \frac {V}{R} = \frac {5}{10k} = 0.5 \ \mathrm{mA} \$ when connected to GND.
If we make a conservative calculation that the transistor has a gain of 100 then we need 5 uA into the base.
The voltage across R1 will be 3.3 - 0.7 = 2.6 V due to the forward voltage drop across the base-emitter junction. Let's use 2.5 V to make the calculation easy.
\$ R1 = \frac {V}{I} = \frac {2.5}{5u} = 500k \$.


Most of us, at this stage, would use a 100k resistor to apply a large safety factor and ensure that the transistor is definitely driven into saturation to ensure that the input is pulled low.