Can someone help me figure out what resistor to wire to this LED for maximum brightness?
According to the Industar Luxdrive datasheet the LED is a Cree XP-E. In that datasheet the red and orange-red versions have this current vs voltage curve:-

So you can expect the LED to drop between ~1.9 V and 2.25 V depending on current draw.
A typical 3.7 V Li-ion battery has an operating voltage range of ~4.1 V to 3.0 V under load. If you choose a maximum current of 500 mA with a fully charged battery the voltage difference will be ~ (4.1 V - 2.2 V) = 1.9 V, and the required resistance is ~ 1.9 V / 0.5 A = 3.8 Ω (the nearest 5% series value of 3.9 Ω will be close enough).  The resistor will dissipate ~ 1.9 V * 0.5 A = 0.95 W, so it should be rated for at least two watts.
The LED itself will consume ~2.2 V * 0.5 A = 1.1 W, which doesn't sound like much. But the light output will probably be roughly equivalent to a 12 W  incandescent bulb with red filter. As the battery runs down the current will drop to ~ (3.0 V - 2.0 V) / 3.8 Ω = ~ 0.26 A, producing about half the light output. Again this sounds worse than it will look, as the human eye is not very sensitive to absolute brightness.