Mm, kind of. But place the zener only across the cap. And you can do with just one resistor for the whole bank.
And yes, the zener should be capable to handle the power. But if you connect them only across the caps, not the resistor, they will max dissipate I x 2,4V = P. So if you charge them with a constant current of 1A it becomes 1A x 2,4V = 2,4W.
But you are NOT charging them with constant current but with a resistor. You calculated a resistor of 5Ohm to limit the charge to 1A max (that's only at the start when the full 5V is over the resistor). So, worst case we have one full cap (aka, zener wants to conduct) and one empty cap (0V over it). This leaves 5V - 2,4V = 2,6V for the resistor. 2,6V / 5Ohm = 520mA. So the zener will only get 520mA so it will only produce 2,4V x 520mA = 1,25W.
And that's worst case! In real live you try to charge the caps the same so when one zener starts to conduct (cap is full) the other cap will be almost full. So let's say the voltage on the other cap is 0,3V lower then the full cap. Then we have 2,4V + 2,1V = 4,5V over both caps. This leave 5V - 4,5V = 500mV for the resistor. 500mV / 5Ohm = 100mA will flow into the caps. The zener will have to burn 2,4V x 100mA = 240mW. So even a halve watt zener will do if you try to balance charge them and you use a resistor of 5Ohm to charge them 