Interesting result. How did you measure voltage and current?
To that comes why the wavetops for the yellow graph is apruptly cut of, compared to the round tops of the orange and the red graph.

The yellow graphs follows the power supply  probably similar to a sine wave (minus some offset for the diode).
The peak values of the voltage increases with the capacitance, so that kind of explains it

No, as the additional current is needed to charge the capacitor. It discharges itself via the lamp when the power supply is at a negative voltage.
One guess: The capacitor increases current and therefore the average load of the lamp  it gets hotter, and its resistance might increase. If the power supply is not perfect and its voltage depends on the current, this might lead to a higher peak voltage. I would not expect this, but something has to explain your observations and I don't see anything better at the moment.