Radical invariance of first nonzero Ext

Since $I$ and $J$ are finitely generated and have equal radicals, some positive powers satisfy $I^a\subseteq J$ and $J^b\subseteq I$: take powers of each generator lying in the other ideal, then a sufficiently high power of the whole ideal lies there. Fix $d\ge 0$ and suppose ${\mathrm{Ext}}_R^i(R\mathrm{/}I,M)=0$ for $0\le i<d$. The same vanishing holds for every finite $R\mathrm{/}I$-module $N$. Indeed, for each such $N$ choose a finite free $R\mathrm{/}I$-module $F$ surjecting onto it, with finite kernel $K$. For $i=0$, ${\mathrm{Hom}}_R(N,M)$ injects into ${\mathrm{Hom}}_R(F,M)=0$. Inductively, for $1\le i<d$, the long exact Ext sequence places ${\mathrm{Ext}}_R^i(N,M)$ between ${\mathrm{Ext}}_R^{i-1}(K,M)=0$ and ${\mathrm{Ext}}_R^i(F,M)=0$. If a finite module $N$ is killed by $I^r$, its finite filtration $N\supseteq IN\supseteq \cdots \supseteq I^{r}N=0$ has finite $R\mathrm{/}I$-module quotients, so the long exact sequences give the same vanishing for $N$. The converse follows by taking $N=R\mathrm{/}I$.

Apply this first to $R\mathrm{/}J$, annihilated by $I^a$, and then symmetrically to $R\mathrm{/}I$. The two Ext families vanish through the same initial range, including the all-vanishing case.