It may be that the constant current this circuit produces isn’t enough to effectively drive the LEDs. It would be interesting to know the voltage drops across the LEDs, the transistor and the resistor when the control line is fixed in the high state.
Theoretically, if we assume the voltage drop across the LEDs is 1.2V each, or 2.4V in total, and the drops across the regular diodes and the BE junction of the transistor are all 0.7V. Then your circuit peggs the Base voltage at 1.4V maximum, which puts the Emitter voltage at 0.7V, which is also the voltage across the resistor. If the resistor is 2.5 ohms and drops 0.7V, then the current must be 280mA. I suspect that’s not enough even with two IR LEDs. Then if the supply voltage is 5V, you are dropping 2.4V across the LEDs, 1.9V across the transistor, and .7V across the resistor.
The attached alternative circuit has everything rearranged, and the constant current feature removed. I haven’t actually tried it, so you would have to see if it works. Anyway, by placing the LEDs at the bottom and the resistor at the top, I think the transistor will be fully saturated, and there will be no voltage drop across it. In other words, there will be 2.4V dropped across the LEDs and 2.6V dropped across the resistor, the total current will be 1.04A. In addition, this arrangement even runs the base current through the LEDs, so nothing is wasted.
By the way, 1 amp may be too much. You might try a 4.7 ohm resistor at first, and see if that’s enough power, or better yet, add a third LED if you have an extra. Either would drop the current to about half an amp.
Edit: You also have to be sure your transistor and whatever’s driving the base are up to the task of handling 1 amp.
Edit2: I forgot to deal with the base resistor. If the transitor has gain of 50, you might need a base resistor as low as 100 ohms to drive 1A through the transistor. If supplying enough base current is a problem, you might have to switch the NPN to a darlington.