Why don't pure solids or liquids shift equilibrium?

Because their concentrations are essentially constant. Concentrations only change easily for things that #(i)# have solute in them and #(ii)# are not rigid.

An equilibrium constant in standard conditions is written for, e.g.

#aA + bB rightleftharpoons cC + dD#

as

#K^@ = (([C]//c^@)^c([D]//c^@)^d)/(([A]//c^@)^a([B]//c^@)^b)#

where these concentrations #[" "]# are equilibrium concentrations. #c^@ = "1 M"# is the standard state concentration for aqueous solutions, i.e. at #25^@ "C"# and #"1 atm"#. This does not apply for pure liquids and solids.

PURE LIQUIDS

For pure liquids... if there is no solute in solvent... it must be a pure liquid and thus it has no true concentration. One might say its solute concentration is #"0 M"#, but clearly that would make #K -> oo# or #K -> 0#, so we do something else instead.

We specify that its standard state is its molar density, #rho = "mols liquid"/"L liquid"#, and this is precisely its actual "concentration" if you calculated its "concentration" as #"mols"# of itself "dissolved" in a volume of itself.

Thus, #(["pure liquid"])/(["pure liquid"]^@) = (["pure liquid"])/(bar rho) = 1# and it essentially does not vary.

PURE SOLIDS

It is even more true for solids. Again, we use the standard state of its molar density, and if you calculated a "concentration", that would match the molar density, i.e.

#(["solid"])/(["solid"]^@) = (["solid"])/(bar rho) = 1#

And it again does not vary.