Help with piezoelectric circuit
The bulb didn't light up even for a fraction of a second.

It probably did illuminate but, it would have done so for a small fraction of a second. For a capacitor, we can say this: -
$$Q = C\cdot V$$
And, if we differentiate charge (Q), we get current and rate of change of voltage: -
$$i = C\cdot\dfrac{dv}{dt}$$
So, if the LED is taking about 30 mA\$^{note1}\$, the rate at which the voltage collapses on the discharging capacitor is this: -
$$\dfrac{dv}{dt} = \dfrac{30\text{ mA}}{100\text{ μF}} = \text{300 volts per second}$$
So, assuming the LED needs at least 2 volts to operate and, the capacitor is charged up initially to 6 volts, then the LED lights for about 13 ms (give or take a few milliseconds).
You could try making the 100 Ω resistor a lot bigger (say 1 kΩ) so that there will only be about 3 mA flowing (on average) into the LED and it should still light but, it will still only light for maybe 130 ms. Turn the lights down to see it.

\$^{note1}\$ - The resistor in series with the LED is 100 Ω hence the initial current surge will be 40 mA, but I have used 30 mA as an approximate to the average level over the short time the LED illuminates. We could argue this one way or the other but, at the end of the day, the illumination duration will still be in the region of 10 to 20 ms.