Yes, the Reddit post's approach is likely to work ok, as long as you correctly follow the cautions in it: • connect the VCC pin of the relay board to a 5V source,
• connect together the Arduino ground, the relay ground, and the 5v-source ground.
Here is a typical relay-module circuit diagram (for one channel):
(Note, this diagram is from a home automation blog; for more discussion and similar diagrams, see eg
4-relay-shield-diagram at yourduino.com and stripboard-shield-relay-sm at startingelectronics.org.)
As you can see in the diagram, the Arduino pin that connects to JP2-2 will not be exposed to voltages. (The base of an NPN transistor effectively is at the junction of two back-to-back diodes.)
When the Arduino or ESP pin at JP2-2 is at about 0 V (ie, when the Arduino pin is off) IBE=0; that is, no Q2 base-emitter current will flow; thus, Q2 will be off, so that both of LED1 and relay U1 will remain off.
When the Arduino or ESP pin at JP2-2 is at about 3.3 V (ie, when the Arduino pin is on) IBE is about 2.5 mA, which should be enough to reduce VCE to about 1 V, leaving 4 V across the relay coil, or enough to turn it on.
Note, the 2.5 mA figure for this circuit is calculated as follows: IBE = (3.3-VBE)/1000 - VBE/10000 = (3.3-0.7)/1000 - 0.7/10000 = .00260 - .00007 = .00253 A = 2.53 mA. Note that the alternate circuit diagrams linked above don't have a 10KΩ resistor like that in the diagram above, so instead of about 2.5 mA would have about 2.6 mA drive.
If for some reason you don't have quite enough base drive – in which case you might be able to hear a slight click from the relay, even if it doesn't switch – you could increase the "5 V" line into the LED and relay to say 5.5 V or 6 V. As noted before, the back-to-back diodes structure of the NPN transistor will isolate that voltage from your Arduino or ESP.