3.3 volts > 1.4 volts. So 3.3 volts should keep it shut off.
But for more confusion see also note 5. The two sentences seem contradictory to me.
EDIT: Here's a pic of what I'm talking about. Push-pull totem output is internal to the microprocessor. You should reduce \$R_{FA}\$ by the output resistance of the totem pole, which is in the vicinity of 25 ohms. It's in the micro's datasheet.
I've designed a small system that might work for your particular application. Here is the schematic:
I forgot the reference node, it must be connected to terminal 2 of BT1.
How does this work?
From high to low
First of all let's assume we can neglect the current flowing in/from "low".
When high is pulled up (5V) in R1 no current flows, while the three diodes are conducting. Assuming a forward drop voltage of 0.6V the voltage at low will be 3.2V, and the current flowing from high will be approximately 320uA.
When high is pulled down (0V) all the diodes are interdicted, so the voltage at low will be pulled down by R2. The current that high must sink is approximately 500uA.
From low to high
Now let's assume high is not consuming current.
When low is pulled up (3.3V) the diodes can not conduct because the voltage at high would be more than 5V, so high is pulled up by R1, the diodes are off and low must provide about 330uA.
When low is pulled down (0V) the diodes are correctly polarized, R2 has zero volt across, the voltage at high is about 1.8V and the current sunk by low is approximately 180uA.
As you can see, the big problem is that 1.8V is a bit too much: a CMOS circuit would probably read that "low", while a TTL is likely to read that "high". A better approach could use a 1.5V zener diode instead of the three small signal diodes, with the cathode connected to R1 and the anode to R2. The resistor will probably need to be reduced to meet the minimum polarization current of the zener diode.
One last thing about the resitors is that you can use any value from 1k to 100k, of course higher resistance values correspond to lower current consumption, but also to slower transient response, and vice versa.
Best Answer
There generally is no need to use a level shifter to control a BJT transistor, as long as the high output level is higher than the required \$V_{\rm BE}\$ (1.2V for your BC337-16 transistor, with \$I_{\rm C} = 300 \rm mA\$ and \$V_{\rm CE} = 1 \rm V\$). WHat are you trying/willing to do?