1) I would give the gate resistor (R14) of Q2 a higher value than 0E (0 ohm). Something like 47E.
2) The nets on both sides of 14 are called "ID-Diode" - that shorts the resistor. Call the one of the MCU "IR-Diode-mcu".
3) The BAT54 (D1) has a high forward voltage drop at high current. I suggest a better diode like PMEG3010ER
https://assets.nexperia.com/document...PMEG3010ER.pdf. Do note, you will need two of the PMEG3010 diodes.
4) With higher current pulses taking power from Vin+, there is only C1 that is 1uF. Looking at the symbol it is a non-polarized type, probably a X5R or X7R. These capacitors are notorious for DC-bias and AC-bias. A better capacitor is this one:
C1206C106K4RACAUTO7210 from Kemet. It has about 20% loss at 5V, so you are left with about 8uF. Alternatively, place this capacitor addinally close to the LED connector and have the MOSFET (Q2) there too.
5) If you go for the 10uF X7R I mentioned in the previous point, you can also use it at the USB power input and replace C3. C3 in the original design is a tantalum capacitor.
6) The current limiting resistor for the LED, R2 will 'burn' the excess voltage into heat. If you want to go to the maximum current in the future (100mA), you may want to use either 2x 1206 in series or parallel, or go for a bigger package 2512. A 1206 resistor is commonly only 250mW. (5V-1.35V)*100mA = 365mW.
7) I see the text 1.27mm at H6, the connector to the LED. I would go for a more standard 2.5 or 2.54mm.
8) Better place a capacitor near H1 (battery connector) of at least 100nF as an ESD protection.