Let $M,N$ be $R$-modules. Since $\operatorname{pd}(M)\leq n$, we have a projective resolution, $$\cdots\rightarrow0\rightarrow P_n\rightarrow\cdots\rightarrow P_0\rightarrow M\rightarrow0.$$
Then $\operatorname{Ext}_{R}^{n+1}(M,N)=\ker d^{n+1*}/\operatorname{im}d^{n*}$ where $$d^{n+1*}:\operatorname{Hom}_R(P_n,N)\rightarrow \operatorname{Hom}_R(P_{n+1},N).$$
Since $P_{n+1}=0$, $\operatorname{Hom}_R(P_{n+1},N)=0$. Then $\ker d^{n+1*}=\operatorname{Hom}_R(P_n,N)$.
Is $\operatorname{im}d^{n*}=\operatorname{Hom}_R(P_n,N)$?
There is a problem with notation. We have a projective resolution $$ 0 \xrightarrow{d^{n+1}} P_n \xrightarrow{d^{n}} P_{n-1} \xrightarrow{d^{n-1}} \cdots \xrightarrow{d^{2}} P_1 \xrightarrow{d^{1}}P_0 \rightarrow M \rightarrow 0 $$ and the $\operatorname{Ext}_R^j(M,N)$ groups are computed as the homology of the complex $$ 0 \rightarrow \operatorname{Hom}(P_0, N) \xrightarrow{{d^{1}}^\ast} \operatorname{Hom}(P_1, N) \xrightarrow{{d^{2}}^\ast} \cdots \xrightarrow{{d^{n}}^\ast} \operatorname{Hom}(P_n, N) \xrightarrow{{d^{n+1}}^\ast} 0 \cdots $$ Then (in this notation) $\operatorname{Ext}_R^{\color{red} j}(M,N) = \ker {d^{\color{red}{ j+1}}}^\ast / \operatorname{im} {d^{\color{red} j}}^\ast$ for every $j\geq 0$. Just need to note that $\operatorname{Ext}_R^{\color{red} 0}(M,N) = \ker {d^{\color{red} 1}}^\ast$.
Clearly $\operatorname{Ext}_R^{n+1}(M,N) = \ker {d^{n+2}}^\ast / \operatorname{im} {d^{n+1}}^\ast= \{0\}$ since the relevant modules are $P_{n+1} = P_{n+2} = \{0\}$.