If the gravitational field contributes to the stress energy tensor, why is $T^{\mu\nu}=0$ in a vacuum solution? I have read that the Einstein Field equations are non linear and gravitational fields can themselves be a source of curvature.
Why do we then while solving for a vacuum solution take $T^{\mu\nu}=0$?
Won’t a non zero gravitational field in vacuum contribute to the stress energy tensor and thereby give rise to curvature itself? In that case and if it is true that the gravitational field itself gives rise to curvature then we shouldn’t take $T^{\mu\nu}=0$ even in vacuum. Why do we that then?
 A: By definition the vacuum means $T_{\mu \nu} = 0$. The energy-momentum tensor is defined as the variation of the matter action $S_{M}$ with respect to the metric, it doesn't include the 'energy' of the gravitational field.
You can consider different energy-momentum tensors (or pseudo tensors) which don't just represent the matter fields, but that's a different issue. E.g. see How is the approximate gravitational wave stress energy momentum tensor not 0? and Energy-Momentum Tensor of a Gravitational Wave)
A: The degrees of freedom of the gravitational field in Einstein gravity are captured in the Riemann curvature tensor $R^\mu{}_{\nu\kappa\lambda}$ which has 20 independent components (in dimension 4). The Einstein equation determines the partial trace $R_{\mu\nu} = R^\lambda{}_{\mu\lambda \nu}$ of the Riemann tensor (known as the Ricci tensor), which has ten independent components itself. That is, ten components of the Riemann tensor are not determined by Einstein equations. The undetermined components are captured by the so-called Weyl curvature tensor $C^\mu{}_{\nu\kappa\lambda}$, which is the completely traceless part of the Riemann tensor and has ten independent components.
That is, there are degrees of freedom of the gravitational field (degrees of freedom of curvature) that are not determined by the Einstein equation! So saying that $T^{\mu\nu} = 0$ and thus $R^{\mu\nu} = 0$ is not equivalent to saying there is no curvature. In other words, Einstein equations tell you that "matter = some aspects of curvature", but a part of the curvature (the Weyl tensor) is not determined by matter.
How do we call space-times that have only $C^\mu{}_{\nu\kappa\lambda}$ and no $R_{\mu\nu}$? We could call them matter-free. However, an observer in free fall sees this space as completely empty in a local inertial frame, there are no fields to be detected! (By the Einstein equivalence principle on which entire general relativity is built.) Well, if what every observer will see is just empty space in this metric, we might as well call it a vacuum space-time.

So, you might ask, how is $C^\mu{}_{\nu\kappa\lambda}$ determined? I would say that this could be called the hard problem of Einstein gravity. Mathematically, the Weyl tensor is determined by boundary conditions. Physically, we would translate boundary conditions as the history and global context of the system in question.
Let us take a look at systems that are in asymptotically flat space-time, that is, they start as isolated and in a flat universe. In these space-times you always need some matter as a seed of the curvature $C^\mu{}_{\nu\kappa\lambda}$. That is, in this context curvature cannot arise on its own, without the seeding of matter. However, once some matter is present, the curvature can self-multiply in the sense that if you take two particles that would gravitate with a certain gravitational mass on their own, they will gravitate with a larger mass when brought together. Typically, vacuum space-times have some matter hidden somewhere implicitly in the boundary conditions, but if $T^{\mu\nu}=0$ everywhere apart from that, we tend to still call these space-times as vacuum.
