Let $(R, \mathfrak m)$ be a one-dimensional commutative Noetherian local ring. Let $M$ be a finitely generated $R$-module with finite injective dimension $\operatorname{injdim}_R(M).$ One can prove that $R$ is Cohen-Macaulay by showing first that $\operatorname{injdim}_R(M) \geq 1$ (cf. this post) and subsequently appealing to the fact (proven by Bass) that $\operatorname{injdim}_R(M) = \operatorname{depth}(R)$; however, my aim is to prove the following.
Let $(R, \mathfrak m, k)$ be a one-dimensional commutative Noetherian local ring. If $R$ admits a finitely generated reflexive $R$-module $M$ with finite injective dimension, then $R$ must be Gorenstein.
Unfortunately, I fear that the line of reasoning used to prove that $R$ is Cohen-Macaulay is of little help in proving that $R$ is Gorenstein. But at the same time, once we know that $R$ is Cohen-Macaulay, we can invoke Grothendieck's Local Duality Theorem. Ultimately, we must show that $\operatorname{Ext}_R^1(k, R) \cong k$ and $\operatorname{Hom}_R(k, R) = 0.$
Edit: As @metalspringpro points out, there was an unresolvable error in my initial attempt, so I have removed it. Furthermore, it suffices to prove that if $R$ admits a reflexive canonical module $\omega_R,$ then $R$ is Gorenstein.
I would greatly appreciate any advice, comments, or suggestions.