Are the output pins on the 74HC595 shift register protected against shorting?
In this data sheet it says:


  When the output-enable (OE) input is high,the outputs are in the
  high-impedance state.


However on page 3 you will notice that the *OE pin controls all outputs at the same time.  So if you connect 2 output pins and one is high and the other is low and you hold *OE low ... you will likely burn out the driver for one of the 2 output pins if not the chip its self.

If you add diodes to BOTH outputs you are essentially creating a wired or gate.  (You should add a resistor.  See the examples in the provided link.) This is fine as long as you understand how a wired or gate works.