The current in the forward bias is known to be more (≈mA) than the current in the reverse bias (≈µA). What is the reason, then, to operate the photodiode in reverse bias?

In a reverse biased p-n junction, the width of the depletion region increases as you increase the applied reverse bias voltage across the diode (proportional to the square root of the voltage). So, by applying a larger voltage, more of the incident photons are converted to electric current, or the efficiency increases. On the other hand, in a forward bias of a p-n junction, the width of the depletion region reduces, so, only a small portion of the incident photons get converted to electric current.