I apologize for the vagueness of this question:
How would one show that a Riemann surface $(X,g),$ endowed with a Riemannian metric $g,$ is conformally equivalent to the complex plane? (i.e. is there a standard program to prove such a result?)
Heuristically, I assume you would need to show it is a Riemann surface of genus one. The Uniformization Theorem would then imply that the universal cover of the Riemann surface is the complex plane.