Your argument is very useful. I follow your argument. Since $R/I$ has finite length, we know $$R/I\cong \mathrm{Hom}_R(\mathrm{Hom}_R(R/I,E_R(k)),E_R(k))=\mathrm{Hom}_R(N,E_R(k)).$$ You have showed that $\mathrm{pd}_RN<\infty$, so we know $\mathrm{id}_R(R/I)<\infty$.
Lemma 1. Let $(R,m,k)$ be a Noetherian local ring. If $M$ is a finitely generated $R$-module, then $\mathrm{id}_RM=\sup\{i\mid \mathrm{Ext}^i_R(k,M)\neq 0\}$.
By lemma 1, and by Nakayama we know:
Lemma 2. Let $(R,m,k)$ be a Noetherian local ring, $M$ is a finitely generated $R$-module, and $x\in m$ is a regular element on $M$. Then $\mathrm{id}_R(M)=\mathrm{id}_R(M/xM)$.
So we know $\mathrm{id}_R(R)=\mathrm{id}_R(R/I)<\infty$, since $I$ is generated by a regular sequence.
More generally, Foxby showed:
If $R$ is a Noetherian local ring, if there exists a finitely generated module with finite injective dimension and finite projective dimension, then $R$ is Gorenstein.
Henselian is superfluous. Here need $(R,\mathfrak m)$ is a Cohen-Macaulay local ring.
$(\Leftarrow)$ In order to prove this, we just need to prove that the second condition implies that $M$ is a $d-$th syzygy of some module for any $d-$dimensional commutative noetherian ring.
First, recall that there is a long exact sequence
$$
0\rightarrow \rm{Ext}^1_R(\rm{tr}(M),R)\rightarrow M\rightarrow M^{\ast\ast}\rightarrow \rm Ext_R^2(\rm tr(M),R)\rightarrow 0.
$$
We may assume that $d\geq 3$. The condition that $\rm Ext_R^i(\rm tr(M),R)=0$ for $1\leq i\leq d$ is equivalent to that $M\cong M^{\ast\ast}$ and $\rm Ext_R^i(\rm tr(M),R)=0$ for $3\leq i\leq d$. Since $\Omega^2_R(\rm \rm tr(M))\cong M^\ast$. Thus $\rm Ext_R^i(\rm tr(M),R)=0$ for $3\leq i\leq d$ is equivalent to $\rm Ext^i_R(M^\ast,R)=0$ for $1\leq d-2$.
Choose a free resolution of $M^\ast$
$$
F_{d-1}\rightarrow \cdots\rightarrow F_1\rightarrow F_0\rightarrow M^\ast\rightarrow 0.
$$
The argument above means that the dual of above exact sequence is exact. That is, the following exact sequence is exact
$$
0\rightarrow M\rightarrow F_0^\ast\rightarrow F_1^\ast\rightarrow \cdots\rightarrow F_{d-1}^\ast.
$$
In particular, $M$ is a $d-$syzygy of an $R$-module.
$(\Rightarrow)$ $d=0$ is OK. For $d=1$, recall that there is an exact sequence
$$
0\rightarrow \rm Ext_R^1(\rm tr(M),R)\rightarrow M\rightarrow M^{\ast\ast}.
$$
Denote $L=\rm Ext^1(\rm tr(M),R)$. Since $M$ is locally free on the punctured spectrum, $\rm Supp L=\{\mathfrak m\}$ if $L\neq 0$. So $L$ is of finite length. Denote the cokernel of the map $L\rightarrow M$ is $L^\prime$. We observe that $M^{\ast\ast}$ is MCM. It follows that $\rm depth(L^\prime)>0$. However, the depth lemma yields that
$$
\rm depth(L)\geq \min\{\rm depth(M),\rm depth(L^\prime)+1\}>0.
$$
This contradicts with that $L$ has finite length. Hence $L=0$.
Similarly, one can prove the case of $d=2$. Then $M$ is reflexive. This yields that $\rm Ext_R^i(\rm tr(M),R)=0$ for $i=1,2$.
Now assume that $d\geq 3$. As before, one can prove $M$ is reflexive ($\iff \rm Ext_R^i(\rm tr(M),R)=0$ for $i=1,2$). Since $\Omega^2_R(\rm tr(M))\cong M^\ast$, it remains to show that $\rm Ext^i_R(M^\ast,R)=0$ for $1\leq i\leq d-2$. Assume hat $\rm Ext^i_R(M^\ast,R)\neq 0$ for some $1\leq i\leq d-2$. Let $j$ be the smallest integer such that $\rm Ext^j_R(M^\ast,R)\neq 0$. Choose a free resolution
$$
P_j\rightarrow P_{j-1}\rightarrow \cdots\rightarrow P_1\rightarrow P_0\rightarrow M^\ast\rightarrow 0.
$$
Dualizing the above resolution, our choice of $j$ yields that we have an exact complex
$$
0\rightarrow M\rightarrow P_0^\ast\rightarrow P_1^\rightarrow \cdots\rightarrow P_{j-1}^\ast\rightarrow P_j^\ast\rightarrow C\rightarrow 0.
$$
We observe that $\rm Ext^j_R(M^\ast,R)$ is a submodule of $C$. Note that $\rm Ext^j_R(M^\ast,R)$ has finite length as $M$ is locally free on the punctured spectrum. Hence $\rm depth(C)=0$. Then the depth lemma yields that $\rm depth(M)=j+1\leq d-1$. This contradicts with $M$ is MCM. Thus we have $\rm Ext_R^i(M^\ast,R)=0$ for $1\leq i\leq d-2$.
Best Answer
Let $R=K[X,Y]/(X^2Y)$ and $M=R/(xy)$. Then $M$ is MCM, $\operatorname{Ann}_R(M)=(xy)$, and $\operatorname{Supp}(M)=\operatorname{Spec}(R)$ since every prime ideal of $R$ contains $x$ or $y$.