Transistor switch: How to commutate 4-6V rail with 0-3.3V signal?
What you are describing is called the "emitter follower" or "source follower" configuration.  That's not what you want here since you want to drive the buzzer with a higher voltage than the 3.3V the microcontroller puts out.

What you want is to use the transistor in common emitter (BJT) or common source (FET) configuration.

For a FET, connect the source to ground, the gate to the microcontroller output, the drain to the low side of the buzzer, and the 4-6V supply to the high side of the buzzer.  In this case the FET needs to turn on well with only 3.3V gate drive.  Those are called "logic level" FETs.  I often use IRLML2502 in such applications.

For a BJT use a NPN with emitter, base, and collector the way the FET source, gate, and drain were hooked up.  Except there needs to be a resistor between the base and the microcontroller output.  Let's say the buzzer draws 100mA and the transistor has a minimum gain of 50 in this case.  That means you need at least 2mA base drive.  Figure the B-E drop is 700mV, so that leaves 2.6V accross the resistor.  2.6V / 2mA = 1.3KOhms.  So 1KOhms would be reasonable to leave some margin for the example values I used.

In either case, make sure to put a reverse diode accross the buzzer.  It will be at least partially inductive, so will produce a large voltage spike when turned off unless you give it a path to flow.  Without the diode the transistor will get fried, and bad voltages could feed back to the microcontroller.