Yes, replacing the fifth diode with a zener would work better, although for the values shown you would probably need a fairly high power zener (>=2W probably) to dissipate the current. 1uF is quite high for a 220V mains capacitive dropper. The person in the video doesn't specify the voltage drop across that LED board, but I count 17 LEDs, so my guess is that board is more likely to be a 5V board meant to be run from a USB, which would have all of the LEDs in parallel and would be very inefficient (having them in series would have a Vf~=51V, which would be too high for the capacitor and I think would have a current of around 50-60mA). Having all the LEDs in parallel would also have very poor mains power factor.
Capacitive droppers are more efficient and have better (relatively) power factor when run at lower current (i.e. lower value on the mains-side film cap) through a higher voltage series string of LEDs. Of course, you would also need to use a higher voltage DC filter capacitor and zener, so some LED bulbs just omit the zener and just use a 400V-rated capacitor (the extra-cheap ones just omit the capacitor entirely and are super flickery). For a 120V (170Vpeak) mains voltage, one could use a lower voltage cap but you would need a higher value on the film cap for the same current.
Quite frankly, this isn't the best or most efficient example of a capacitive dropper circuit, so you may want to re-work the values if you plan to build one.