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Generally the output pin of the MCU will be, between 4.3V and 5V when high and will allow a maximum current of 15mA, an LED will have a volt drop of about 2 volt in forward bias conditions at 10mA which is about the minimum current to make it glow normally.

Place a resistor in series with the MCU pin and the LED, connect the cathode pin of the LED to the ground. As an on state is created on the MCU, the current will flow through the resistor to the LED, through the anode, flow through the LED to the ground

The minimum current for LED at the minimum voltage is must be used to obtain the Rs

Rs=(Vmcumin-Vled)/Imin

Rs=(4.3V-2V)/0.01A

= 230R using a higer value 330R should be fine

Imcu = (5V- 2V) / 330R = 9.1mA it should be safe for your MCU and it should produce a visible indication.

I will not recommend making it to bright, LEDs can easily go to about 20mA but for most applications around 10mA is enough

If one would need to use 15mA or more, then the help of an external transistor is recommended. Then you have VCC going through resistor Rs through LED Anode to Cathode to the collector of a NPN BJT transistor. Then from the Emitter to the ground. From the MCU let the voltage go though resistor Ri to the Base of the transistor. Generally it is bad to switch loads this way, since the loads are never properly grounded and said to be floating, but for LED it is ideal.

Then

Rs = (VCC-Vled-Vce) / Iled = (5V-2V-0.2V)/0.01A = 280R NAV = 330R

Ic = (Vcc-Vled-Vce)/Rs = (5V-2V-0.2V)/330R = 8.47mA

then

ib = Ic/Beta = 8.47/100 = 84.8uA

to allow some tolerance let Ibnew = Ib x 10 = 848uA

Ri = (Vmcumin - Vbe)/Ibnew = (4.3V-0.7V)/848uA = 4.245K NAV = 4k7

The transistor:

Ie= (Vmcumax-Vbe)/Ri = (5V-0.7V)/4.7k = 915uA

Power dissipation for transistor is

Pd= Vce x Ie = 0.2V x (Ib + Ic) =0.2 x 0.009395 = 1.88mW

one can see, a very small transistor is required.

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