I am posting my comment as an answer. Perhaps the result that you are looking for is Ischebeck's Theorem. <B>Ischebeck's Theorem.</B> For finitely generated modules $F$ and $G$ over a Noetherian local ring $(R,\mathfrak{m})$, $\text{Ext}_R^i(F,G)$ is the zero module for all $i < \text{depth}(G)-\text{dim}(\text{supp}(F))$. Thus, for instance, if $G$ is a Cohen-Macaulay module of $\text{depth}(G)=\text{dim}(\text{supp}(G)) = d$, then it is true that $\text{Ext}^i_R(F,G)$ is the zero module for $i< \text{dim}(\text{supp}(G))-\text{dim}(\text{supp}(F))$. However, this can easily fail if $G$ is a module with small depth. For instance, if $(R,\mathfrak{m})$ is a Noetherian local ring of depth $d\geq 2$, then for $i < d$, $\text{Ext}^i_R(R/\mathfrak{m},R)$ is the zero module. Moreover, if $R$ is Gorenstein, then $\text{Ext}^d_R(R/\mathfrak{m},R)\to \text{Ext}^d_R(R/\mathfrak{m},R/\mathfrak{m})$ is injective. Thus, using the long exact sequence of Ext modules associated to the short exact sequence, $$\Sigma: \ \ \ \ \ 0\to \mathfrak{m} \to R \to R/\mathfrak{m} \to 0,$$ it follows that the connecting map $$\delta_\Sigma^i:\text{Ext}^i_R(R/\mathfrak{m},R/\mathfrak{m}) \to \text{Ext}^{i+1}_R(R/\mathfrak{m},\mathfrak{m}),$$ is an isomorphism for $i\leq d-2$, resp. for $i\leq d-1$ if $R$ is Gorenstein. If $R$ is a regular local ring of dimension $d$, then each $R$-module $\text{Ext}^i_R(R/\mathfrak{m},R/\mathfrak{m})$ is a vector space over $R/\mathfrak{m}$ of dimension $\binom{d}{i}$. So, in this case, for every $i$ with $1\leq j \leq d$, $\text{Ext}_R^j(R/\mathfrak{m},\mathfrak{m})$ is nonzero. In particular, the depth of $\mathfrak{m}$ equals $1$. Presumably the reason for the hypothesis that $2\leq i$ is the evident nonzero element of $\text{Ext}^1_R(R/\mathfrak{m},\mathfrak{m})$ coming from the non-split short exact sequence $\Sigma$. However, even in the "best" case of regular local rings, $\text{Ext}_R^j(R/\mathfrak{m},G)$ can easily be nonzero for $R$-modules $G$ that have support of large dimension yet have small depth.