Suppose $R$ is a local Artinian algebra.
Question: Is $R$ an injective $R$-module?
Consider a field $k$ and the $4$-dimensional $k$-algebra $A$ freely generated by $x$ and $y$ such that $x^2=y^2=yx=0$. The indecomposable injective module has a two dimensional top (this is, more or less, the socle of $A$), yet $A$ has a simple top because it is local.
In fact, most local algebras are not self-injective.
No. Here is a commutative example. Take a field $k$ and the 3-dimensional (hence artinian) $k$-algebra $R=k[X,Y]/(X^2,XY, Y^2)=k[x,y]$. This algebra is not injective over itself: here is why.
Consider the ideal $I=Rx=kx\subset R$ . The map $f:I\to R:\lambda x \mapsto \lambda y \quad (\lambda \in k)$ is $R$-linear, but cannot be extended $R$-linearly to $R$ and thus $R$ is not $R$-injective. Indeed if $F:R\to R$ were such an extension, then we would get $y=f(x)=F(x)=xF(1)$. But this is absurd because $y$ is not a multiple of $x$ in $R$, that is $y \notin Rx=kx$.
I don't know if this will interest you or not, but I figured you might find it interesting to read at least once.
A ring which is self-injective on a side and Artinian on a side is quasi-Frobenius, and it actually turns out it is Artinian and self-injective on both sides.
Among commutative local rings, the quasi-Frobenius (=self-injective) ones are exactly those which are Artinian and which have one and only one minimal nonzero ideal contained in all the other ideals.
The same cannot be exactly said for noncommutative rings, but something similar holds which is too large to fit in this margin.
You might like to survey the socles of the given counterexamples. To test this out.