So, I found a datasheet here:
http://www.100y.com.tw/pdf_file/SY-IR10,12,15,33,53,74.pdfAccording to the datasheet, it says the typical voltage drop is 1.6v at 20mA. Thus, if you want to use a 4.5v battery, you have to dissipate 2.9v at 20mA. This will give you a resistor of 145 ohms (voltage across the resistor = current through the resistor x resistance. 1 V = 1 A * 1 ohm)
If you want to go brighter, you can, but you'll have to be careful. The limit you're likely to hit first is the 100mW power rating. LED Power = (voltage across LED x current through the LED. 1 W = 1 V * 1 A)
Say you try a 50 ohm resistor. (At this point I'm making up numbers) Then measure the voltage across the LED with a voltage meter to get 2v (the voltage drop slightly increases as current increases.) Then measure across the resistor to find 2.5v (Or measure the battery voltage at 4.5v, then 2v + 2.5v = 4.5v) Then the current is resistor voltage / resistance = 2.5 v / 50 ohm = 50 mA. Then the power dissipated in the LED is 50 mA * 2v = 100 mW, which is the maximum power allowed. So, you can decrease your resistor as long as you don't exceed 100mW (which you'll have to measure.) I recommend giving some safety room, and if I had to guess, I'd say 100 ohm would be decent.
Good luck.
