Projective dimension at most n iff higher Ext vanishes

Assume $\mathrm{pd}(M)\le n$. If $n=0$, then $M$ is projective and positive Ext vanishes by [L3]. If $n\ge 1$, [L1] makes the $n$th syzygy projective, and repeated [L2] identifies every ${\mathrm{Ext}}^k(M,N)$ with $k>n$ with a positive-degree Ext group out of that syzygy, which vanishes by [L3]. Thus (1) implies (2), and (2) implies (3) by taking $k=n+1$.

Conversely, assume (3). For $n\ge 1$, [L2] gives ${\mathrm{Ext}}^1({\mathrm{\Omega }}^{n}M,N)=0$ for every $N$; for $n=0$, read ${\mathrm{\Omega }}^{0}M$ here as $M$. To see directly that an object $X$ with ${\mathrm{Ext}}^1(X,-)=0$ is projective, take the first stage $0\to K\stackrel{i}{\to }{P}_0\to X\to 0$ of a projective resolution and factor $d_1:P_1\to P_0$ as $P_1\stackrel{\pi }{\to }K\stackrel{i}{\to }{P}_0$. The map $\pi$ is a cocycle representing the identity of $K$ in the usual cokernel description of ${\mathrm{Ext}}^1(X,K)$. By [L5] and the assumed vanishing, $\pi =r{d}_1=ri\pi$ for some $r:P_0\to K$. Since $\pi$ is epic, $ri=1_K$, so the sequence splits and [L4] makes $X$ projective. Apply this to $X={\mathrm{\Omega }}^{n}M$ (or $X=M$ when $n=0$), and use [L1] when $n\ge 1$, to prove $\mathrm{pd}(M)\le n$.