Indeed it is unlikely you will find such a high voltage driver with lower than 300mA current, so there are five different alternative ways I would attack this, meaning I'd choose any ONE of them, not a combination of all.
1) If I am to trust your specs for mA and V, then they must all be wired in series. You could cut the PCB traces or wires, so that instead of one series of 14, you instead have 4 parallel series of 3. Yes this only equals 12, you would leave two LEDs out of the circuit, unused.
With 4 parallel, you could run a common 10W driver which is, say 800mA actual (they usually fudge on specs and are a little lower than claimed), divided among the 4 parallel series, is 200mA drive current to each for less than 2/3rds brightness since you're running 12 LEDs instead of 14.
2) The other option is disassemble the included driver and reverse engineer it. Typically you just get the driver IC chip markings and find the datasheet which provides an example circuit for using it in Buck Mode. Typically, setting the current involves one or more adjustment resistors and the datasheet usually includes a formula to calculate the resistor values needed for a particular drive current. Usually, the lower the resistance the higher the drive current. Sometimes such drivers even have two spots on the PCB so you can use a combination of two resistors to arrive at a more precise drive current.
In that case, you could desolder one or the other of the two driver current setting resistors to lower drive current. If there is only one resistor or removal of one of two is an unacceptable light output still, then you would buy the resistor value you need at an electronics supply house like digikey, after having measured the size needed with a caliper if surface mounted.
3) Presumably you like the enclosure. Scrap the LEDs and driver and put whatever LEDs and driver in the enclosure that will give you the lower level of light that you want.
4) Accept some efficiency loss to do it the easy way. Nobody will like this option but it's easy. The driver is set to regulate to 300mA. If you put a power resistor in parallel with the series of LEDs, it will divert whatever % of current you want it to, based on the value of resistor you choose.
For example with a roughly 45V drive current (14 LEDs Vf in series), if you want to cut the drive current in half, that's a loss of 150mA so 45V / 0.15A = 300 Ohms resistor in parallel. Since that also means the resistor has to deal with half the power lost as heat, almost 7W (45V * 0.15A), you should choose at least a 15W resistor and suitably heatsink it, which is going to mean a big metal plate or at least a heat spreader if the light chassis has enough metal to bolt a spreader to it, and you'd want to do that far enough away from any plastic such that it doesn't melt or get brittle.
5) Find out the voltage range of the output on the driver. I mean the REAL output it's capable of through testing. For example right now the light has 14 LED in series, but it is possible that you can just rewire the PCB to omit some of the LEDs in the series, or use a switch to take some LEDs out of that series occasionally when you want it dimmer, such that it ends up only running, say 12 in series, or 10, 9, 8, 7, whatever the # of LEDs in series it takes to get the light level you want, remembering that at some point with too few in series it may not remain stable or might shut down, so during testing you'd want to be monitoring voltage and current. A similar alternative would be to use two different drivers with one powering fewer LEDs in series then you use a dual throw AC input switch to select which driver is running and put isolation diodes on the output of each so the one unpowered isn't receiving reverse current from the one that is powered.
Now the easiest answer. If all of the above are more than you can or at least want to do, buy some other light that's dimmer. :grin2:
