The right diode and the resistor are not needed.
Just omit the resistor and connect the right lamp (L2) directly to the upper switch (S2).
The diode prevents the left lamp being powered when the right lamp is lit.
As others have stated the diode needs to be rated for current taken by the lamp. The diode may get warm in use as it will dissipate some power. A silicon diode would drop about 0.7V so with a 21W lamp taking just under 2A it will create just over 1W of heat. An overrated diode with 5A or 10A capability may be appropriate. A schottky diode would reduce the voltage drop and so produce less heat.
simulate this circuit – Schematic created using CircuitLab
PS. It makes it easier to answer the question if you put reference designators (eg D1, D2, S1, S2) on the various components so the answers can be unambiguous.