Does a photon require an EM field to exist? An excitation in the electromagnetic field is called an electromagnetic wave. The smallest possible part of the electromagnetic wave is called a photon.
But if the photon is always accompanied by an electromagnetic field, then how is the compton effect possible, even theoretically, since if the electron is interacting with the photon, then it is inevitably also interacting with an electromagnetic field?
 A: The electromagnetic field is a quantum field that can manifest itself in different ways, depending on the state of the system. A photon is one manifestation of the electromagnetic field. An ordinary radio wave (produced by an oscillating current in an antenna) is another manifestation of the quantum electromagnetic field. 
It's not quite correct to say that a photon is accompanied by an electromagnetic field; instead, a photon is a manifestation of the electromagnetic field. All of the photons in the universe, all of the radio waves in the universe, and all of the magnetic fields of all of the refrigerator-magnets in the universe are all manifestations of the one-and-only quantum electromagnetic field. 
This can be illustrated explicitly in a simplified version of quantum electrodynamics in which matter (electric currents, etc) is treated as a prescribed source rather than a dynamic entity, so that the only dynamic entity is the quantum electromagnetic field itself. In this model, exact calculations are possible, and we can verify explicitly that the thing we experience as an ordinary macroscopic electromagnetic wave is equivalent to a quantum superposition of many different numbers of photons. These are not two different things; they are two different ways of describing the same thing. The reason an ordinary electromagnetic wave doesn't seem "quantum" is — ironically — closely related to the fact that it can be expressed as a quantum superposition of many different numbers of photons. (This is one of those cases where the math is much more clear than the words!) 

...if the electron is interacting with the photon, then it is inevitably also interacting with an electromagnetic field?

The interaction of the electromagnetic with itself (such as the interaction of photons with the electric field of an electron) is extremely weak. In the simplified model mentioned in the preceding paragraph, it is exactly zero. In the more realistic version of quantum electrodynamics, which includes things like dynamic electrons, the interaction of the electromagnetic with itself is not exactly zero, but it is very weak. Under normal circumstances, it is utterly negligible, although it can become significant at very high energies.
The degree to which the electromagnetic interacts with itself can be quantified by considering the cross-section for photon-photon scattering. This is analyzed in these papers:


*

*"Photon-photon scattering: a tutorial", https://arxiv.org/abs/1111.6126 

*"Photon-Photon Collisions — Past and Future", https://www.slac.stanford.edu/pubs/slacpubs/11500/slac-pub-11581.pdf
According to section 7-3-1 in Itzykson and Zuber's book Quantum Field Theory, the cross-section for photon-photon scattering is 
$$
\sigma \sim \frac{\alpha^4}{m^2}\,\left(\frac{\omega}{m}\right)^6,
\tag{1}
$$
 where $\alpha\sim 1/137$ is the fine-structure constant, $m$ is the mass of an electron, and $\omega$ is the photon's energy. (This expression uses units in which Planck's constant and the speed of light are both equal to 1.) For comparison, according to section 5-2-1 in the same book, the cross-section for Compton scattering is
$$
\sigma \sim \frac{\alpha^2}{m^2}.
\tag{2}
$$
So the cross-section for photon-photon scattering is smaller than the cross-section for photon-electron scattering by a factor of
$$
 \sim \alpha^2\,\left(\frac{\omega}{m}\right)^6.
$$
Quantitatively, the electron mass is $\sim$ 500 keV and the energy of an X-ray photon is $\sim$ 100 keV (at the upper end of the X-ray energy range), so photon-photon scattering is weaker than photon-electron scattering by a factor
$$
 \sim \alpha^2\,\left(\frac{\omega}{m}\right)^6
 \sim \frac{1}{137^2}\,\frac{1}{5^6}\sim 10^{-9}
$$
at the highest X-ray energies.
This is one reason why the interaction between photons and the electron's electric field can safely be neglected when considering phenomena like the Compton effect.
