Why it is dimmer on 3.3 V

Where the volts go

The bar is the pin's voltage, split between the two parts in the loop. The blue part is the LED's share, and it barely moves: about 3.0 V, whether the pin gives 3.3 V or 5 V. The orange part is the resistor's share, and it is everything left over.

Only the resistor's share sets the current. From 5 V it is 2.0 V, and 2.0 V across 150 Ω is about 13 mA. From 3.3 V it is 0.3 V, which is about 2 mA.

A third less voltage, a sixth of the current

That is the surprise, and the reason is the fixed share. The pin voltage drops by 1.7 V, and all 1.7 V of the drop comes out of the resistor's share, because the LED's share does not move. 2.0 V becomes 0.3 V, so the current falls to a sixth.

Every LED behaves like this, but a blue one is the extreme case. A red LED takes about 2 V, which leaves a 3.3 V pin more than a volt of headroom. A blue LED takes about 3 V, so on a 3.3 V board it is running on almost nothing.

The 3.0 V is a bench measurement of one board, not a datasheet figure. Your LED's forward voltage may be a tenth of a volt either side of it. At 3.3 V that changes the current a lot, which is why the LED on two ESP32 boards can look slightly different while it looks the same on two Unos.

It still looks lit

About 2 mA is plenty to see indoors. The eye's response to light is closer to logarithmic than linear, so a sixth of the current reads as dimmer rather than as a sixth of the brightness. On a 3.3 V board the LED is a clear indicator, just not a torch.

Making it brighter on a 3.3 V board

The pin cannot give more voltage, and the resistor is soldered to the board, so the honest options are outside it:

• Use a 5 V board. The same block on an Uno gets about 13 mA.
• Switch 5 V with a transistor. The 3.3 V pin drives the transistor, and the transistor connects the LED to 5 V. A transistor as a switch covers the circuit.

Do not try to feed 5 V into SIGNAL alongside a 3.3 V pin, or join the two. The block would be fine; the ESP32's pin would not.