I think that your circuit would be massively improved with constant current circuits feeding the LEDs. The LEDs are 2.2 V, and the supercapacitor get to around 5 V, so there is quite a lot of energy to be obtained between 5 V and 2.2 V, if something more sophisticated than a resistor is used.
Something like the first circuit on
https://www.eecs.tufts.edu/~dsculley/tutorial/opamps/opamps7.html
The voltage reference would have to be separate from the supply, and the op-amp would have to work to low voltages. There would have to be one of these circuits for each LED.
Alternatively, there are boost circuits for LEDs that work very well, and are more efficient. That would put all the LEDs in series, so only one boost circuit would be needed.
A quick search found this
https://www.linear.com/product/LT1937
If you have a 5 F capacitor, charged to 4.8 V, to discharge to 2.5 V the energy available is 41.975 Joules. The LEDs take 5 * 0.02 * 2.2 = 0.22 W, so it will take them 191 seconds to use the 41.975 J of energy. If the switch mode converter is 84% efficient like it says it is, that reduces the time to 160 seconds, so still over 2.5 minutes, which is probably all you need.
I haven't read the data sheet for the boost converter, and there may be limitations that mean the design isn't easy.
I chose 2.5 V as the boost converter won't go lower. There is not much difference in energy available if you go to a lower voltage. The energy is 1/2CV^2, so there is little energy at low voltages. Your post mentioned going down to 0.9 V. Well taking the 5 F capacitor down to 0.9 V would give you 55.575 J instead of the 41.975 J if you go to 2.5 V, which is a significant advantage, but probably a lot of work. However, if you can ditch the Schottky diode, and charge the capacitor to 5.25 V, that would give you an additional 11 J of energy.
If you are using non-rechargeable batteries, there is a lot to be gained by using them down to as low a voltage as possible, to avoid buying batteries more often. However, in this case you are paying for conversion circuitry and design time, when you might be better buying a larger supercapacitor to store more energy.
As I said in the previous post, buy the whole thing. It is much easier. However, if you do want to build something, I think that a boost converter is the way to go. There is no point in boosting to a higher voltage and then using a buck regulator to reduce the voltage, when you can put the LEDs in series for a higher voltage, and simply arrange the boost converter to give out the LED current you want, rather than an intermediate voltage that will need converting again.