Pair production is not the same as decay of a particle. A particle can decay into two components according to its decay probability without needing an extra interaction. A lambda in its rest frame will decay into a proton and a pion, for example, within a predictable decay time .
There is no rest frame for the photon since its mass is $0$ and it is always travelling with the velocity of light. If it were to decay spontaneously into an electron positron pair, they do have a rest mass and a rest frame, and their invariant mass would be at least $2*m_e$, which should have been the mass of the photon. A contradiction.
It can interact though with the fields of other particles . How does the photon interact?
The interaction probabilities can be calculated given the charges of the target particles, the easiest way using Feynman diagrams. One can envisage a photon as sequentially turning into virtual loops of $e^+ e^-$ . One of the virtual electrons interacts with the field of a real charged particle exchanging enough energy and momentum so that both $e+$ and $e-$ become real while energy and momentum are conserved in a three body interaction.
The nucleus helps by ensuring the momentum in the final state ($e^+ e^-$ Nucleus) to be the same as the one in the initial state (photon nucleus).