Your calculations based on the Vz of the zener are a little bit dependant on the zener current. Reverse biased zeners have their specified knee voltage at a specied current, your schematic allows for very small "leakage current", but this also influences the biasing of the zener. This means some experimentation or graph-hunting may be required to find your value.
I would propose a little math to find the best solution to your exact situation with lowest losses:
You want the LED to turn on at about 7.2V. I am going to make the assumption you want it to work in the range of 6V to 7.2V, so we'll use those two as the extremes of our calculation. Further, for ease of the example I am going to assume a hfe of the BC547 of 100. a BC547C type may go up to 500 in the right circumstances.
First off, you mention a low-power LED. This again drives an assumption, as no specifications are made, I will assume the LED current is 2mA and that the red LED adheres to commonalities, needing about 1.9V at that current. Here we go!:
Lowest V+ = 6V, I(led) = 0.002A, V(led) = 1.9V, means: R(led) = (6V - 1.9V) / 0.002A = 2.1k Ohm.
LED current at 7.2V: I(led-7.2) = (7.2V - 1.9V) / 2.1kOhm = 2.5mA
The current into the base of the LED's transistor then becomes 25uA at 7.2V and 20uA at 6V (divided by hfe, which is assumed 100). Which means:
Bias Resistor value, maximum: Rbias = (V+ - Vbase) / I(base) = (7.2V - 0.7V) / 25uA = 260kOhm; or (6V - 0.7V)/20uA = 265kOhm. Meaning 260kOhm is our upper limit. Let's choose 133kOhm to stay well on the safe side:
R(pull-up) = 100kOhm and R(base) = 33kOhm (makes 133kOhm together).
Now, the first transistor is only switched on when the battery is between 9V and 7.2V to keep the LED transistor from turning on. We use that to do the remaining calculations:
I(pull-up) = 9V / 100kOhm = 90uA; or 7.2V / 100kOhm = 72uA. Both values are below 100uA and we want to put at least a couple of uA into the base of the first transistor, so we'll use 1uA as a minimum.
If we want the transistor to turn on at the first sign of a 0.7V difference, we need to assume a differential voltage of 15 to 30mV over the resistor, making:
R(base) = 30mV / 1uA = 30kOhm.
The rest is determined by your choice of Zener and Resistor to determine the initial trip voltage, in this case your transistor set-up will be using about 95uA to 150uA of current when turned off. The rest of the "waste" will be determined by your zener. Once the battery is at the trip level, the zener will start to cut off, so its loss will slowly minimize, but then, of course your LED will be draining 2mA.
For a reliable operation you want Iz not to be too far from its optimum, i.e. not a factor of 100. But there are probably types that can give decent effects with a bias current of 0.5mA.
The schematic becomes:
simulate this circuit – Schematic created using CircuitLab