In this article of the Wikipedia the full photon nucleus interaction for pair production is given
In pair production, a photon creates an electron positron pair. In the process of photons scattering in air (e.g. in lightning discharges), the most important interaction is the scattering of photons at the nuclei of atoms or molecules. The full quantum mechanical process of pair production can be described by the quadruply differential cross section given here
It is not a simple matter.
So, my question is if the photon has a very high energy, lets say 70 MeV or something. Why it does not produce 2 electron and 2 positrons (2 pair or 4 body decay) instead of only 1 pair?
There will be a probability for this to happen, equally complicated as the one in the link , appresicably smaller because there will be more electromagnetic vertices.
For high enough energy the strong interaction takes over , and hadron pairs can be produced, see this for example.