Incidentally there is a way to measure if a 3V3 chip/module is likely to tolerate 5V inputs:
Normally all CMOS inputs have protection diodes to guard against stray static electricity taking the pin outside the voltage range it can handle - these diodes clamp the input in the range -0.5V to Vdd+0.5V approx, where Vdd is the supply voltage to the chip. If you connect 5V via a highish value resistor (10k to 22k sort of range) to the input pin, the diode will clamp the voltage at the pin to about 3.8V if there is a diode.
If there is no diode the pin voltage will be 5.0V... If so this could be because the chip is designed to support 5V inputs. But if not you may have already taken it past its limits.
Protection diodes are small and easy to burn-out, hence the 10k resistor to limit the current.
If you connect a 5V logic output directly to a diode-protected 3V3 input then the output stage of the 5V chip will try to push lots of current through the protection diode, which could lead to failure/deterioration of the diode and/or the output stage.
Putting a 10k resistor in series will act as a rough-and-ready level shifter (in conjunction with the protection diode). the relatively high resistance means somewhat slower edges, sometimes this might matter (fast SPI for instance).