Let $R$ be a finite local ring (with identity) with exactly one minimal left ideal. Is it necessarily true that $R$ has exactly one minimal right ideal !?

The answer seems to be yes. In fact, the unique minimal left $R$-ideal is also the unique minimal right $R$-ideal.

Indeed, let $M$ be the unique maximal ideal of $R$. Then we have a sequence $$ 0=M^{r+1}\subsetneq M^r \subsetneq \ldots \subsetneq M^2 \subsetneq M \subsetneq R $$ of two-sided ideals.

Let $L$ be the unique minimal left ideal. Then $L= M^r$. Indeed, since $L$ is minimal, we have that $ML = 0$ by Nakayama's Lemma, so $L$ is contained in $M^r$, and equality follows from $L$'s uniqueness. Moreover, $L$ is a one-dimensional $R/M$-vector space (otherwise it would not be a minimal left $R$-ideal); thus $M^{r}$ is also one-dimensional.

Now, let $P$ be a minimal right ideal of $R$. Then $P \subset M^{r}$, since $PM=0$ by Nakayama's Lemma. Minimality implies that $P$ is a one-dimensional $R/M$-vector space. By equality of dimensions, we have that $R= M^r = L$.