The circuits and components you propose are rather useless and will in fact not work properly.
The diode will make a voltage drop always, and since it is a regular diode, it will always drop about 0.7V and waste 2 watts at 3A. The RPi would see constant undervoltage of about 4.3V.
Also the 5.1V or 5.6V zeners are not ideal devices and they would start to conduct much earlier than their rated voltage, and will draw more current as the voltage rises. If they are required to clamp a lot of current then they will draw a lot of current before they are supposed to clamp.
According to Model 4B schematics, the 5V node on the GPIO connector is the exact same node that goes directly to the Type-C connector, so it will have exact same protection no matter if you power it via Type-C or via the GPIO header.
Which means, no overcurrent protection anyway. There already is a TVS diode intended to protect from overvoltage and reverse polarity (as it conducts in reverse, hopefully making the source to limit current or burn a fuse).
The difference is between the balance of trying to protect the supply from whatever you connect to it, or trying to protect the RPi from the supply, or trying to protect the RPi from what you might accidentally plug into it so a lot of current runs through it.
The good thing about those mobile phone type supplies is that they generally detect overcurrent or short circuits and shut down due to overcurrent or undervoltage. Same might not be true if you use something more powerful.
