Why do high energy photons produce particle pairs instead of undergoing Compton Scattering?

When an photon with wavelength less then 1.213 pm hit’s matter, instead of undergoing Compton scattering, it disappears completely and is replaced by an electron and positron. This is possible since the incident photon has enough energy to account for the rest energy of the two particles. The likelihood of this happening rather than scattering increases with increased photon energy.

Why do high energy photon-matter interaction prefer pair production at all?

The Compton scattering of photons in material of atomic number $$Z$$ is proportional to $$Z$$, whereas the pair production is proportional to $$Z^2$$.
Hydrogen excluded -- (normally these experiments are carried out at solid material at room temperature which excludes hydrogen) the pair production process is preferred, because is more probable since $$Z^2 > Z$$ for $$Z>1$$.
One could wonder why Compton scattering only goes with $$Z$$: in material of atomic number $$Z$$ the photon can be scattered at $$Z$$ electrons of the shell of the atoms of this material, but this happens incoherently, so therefore the cross section only goes with $$Z$$.
However, when it comes to pair production, one needs another photon which can actually be provided virtually by a electromagnetic field of a nucleus of charge $$Ze$$.
If the probability amplitude $$\cal{M}$$ for this process is computed, the electromagnetic field of the nucleus of $$Ze$$ enters linearly in to the formal expression like $$\frac{Ze}{r}$$ in position space -- or if the computation is done in momentum space the Fourier transform of $$\frac{Ze}{r}$$. However, the cross section of the process behaves like $$d\sigma \sim |\cal{M}|^2 d\Omega$$, so therefore the cross section is proportional to $$Z^2$$.