(Note: I'm assuming $R$ is commutative)
The answer by rschwieb shows why projectives are injective. As to why injectives are projective, note that every injective is a direct sum of indecomposable injectives, which are precisely of the form $E(R/p)$ for prime ideals $p$ of $R$ (as $R$ is Noetherian). It thus suffices to show that $E(R/p)$ is projective for all primes $p$.
Now, $R$ self-injective implies $R = E(R)$, i.e. $\text{injdim}_R R = 0$, which in turn implies that $R$ is Artinian. Thus every prime is maximal, hence minimal, hence associated, so $R/p \hookrightarrow R$ for every prime $p$. Thus $E(R/p) \hookrightarrow E(R) = R \implies E(R/p)$ is a direct summand of $R \implies E(R/p)$ is projective.
(Note: these rings are precisely the Artinian Gorenstein rings, and appear often in contexts of duality).