I know that every compact Riemann surface is algebraic, i.e. every compact Riemann surface is isomorphic to a smooth projective complex curve, not necessarily plane (take e.g. the twisted cubic).
However, I have encountered the following proposition in Freitag's "Complex analysis 2":
Every connected compact Riemann surface is biholomorphically equivalent to the Riemann surface which is associated with an irreducible polynomial $P(z,w)\in \mathbb{C}[z,w]$
This proposition confuses me deeply: what about Riemann surface associated with the twisted cubic? Is it biholomorphically equivalent to the R.S. of a plane curve? How can it be so?
Take a look at the answer to this question: Is every algebraic curve birational to a planar curve.
It's a fairly short argument that every curve is birational to a planar one. Since birational maps between smooth curves are isomorphisms, I think this answers your question. It seems that the statement "not necessarily planar" is false in your statement. The reason that the twisted cubic is not a counterexample is because it is the image of an embedding $\mathbb P^1\to \mathbb P^3$ and hence is isomorphic to a line in the plane.