Any number of ways. 2 D cells is 120mm, so you can easily fit 3 2/3A-size cells lengthwise. or 4 if you go into the tailcap. So 6x17500 in a tri-bore would work; 8x14500 in a quad-bore; even 4x18350 single-stack in an unbored light, although not much runtime that way. I believe I've seen 4x18650 in a 2D quad-bore, but I'm not 100% sure on that.
If you prefer NiMH, you'll need 12 cells (well, 11, but you can fit 12 almost any way you can fit 11), so either 12 1/2A in a tri-bore, or 12 2/3AA in a quad-bore.
If you like primaries, I guess you should just be able to fit 6 CR123s in a two-stack arrangement in an unbored light. Never tried that, though.
I haven't investigated discharge rates for any of these, but it shouldn't be an issue for any of them but the 4x18350 (probably need IMR or IFR for those)-- but make sure you do the math before you commit!
EDIT: I see Linger suggests that 8xAA can't take it. Here's my reasoning:
Driver is buck, efficiency presumed to be 80%.
LEDs are P7 in 3s, Vf=3.5V, so V=10.5V I=2.8A, P=29.4W
Power from battery=P/0.8=37W
Power per cell=Pbat/8=4.625W
Current through cell=Pcell/3.7V=1.25A
14500 capacity = 750mA, max safe current=2C/h = 1.5A, we're drawing less, so OK. I guess it might be a bit marginal as the batteries fall to 3V or so and the current correspondingly rises, but still should be OK.
As for 3x26500, I don't know what overhead the Der Wichtel driver has, but I guess when the OP said "13-14V minimum", he didn't mean 12.6V fresh out of the charger and falling off from there would be fine...