Well, how do we conclude that our complex $Z^\bullet$ is quasi-isomorfic to the complex

$$
\cdots \longrightarrow 0 \longrightarrow Coker(d^{-n-1}) \longrightarrow Z^{-n+1} \longrightarrow \cdots \longrightarrow Z^{m-1} \longrightarrow Ker(d^m) \longrightarrow 0 \longrightarrow \cdots \ ?
$$

We use the two truncations, which are quasi-isomorphisms, due to the vanishing of cohomology with sufficiently small and sufficiently large numbers: $Z^\bullet \longleftarrow \tau_{\le m}Z^\bullet \longrightarrow \tau_{\ge -n}\tau_{\le m}Z^\bullet$.

We can use the stupid truncation $\sigma_{\le -n}$ instead to get the complex

$$
\cdots \longrightarrow Z^{-n-2} \longrightarrow Z^{-n-1} \longrightarrow Z^{-n} \longrightarrow 0 .
$$

Its only cohomology lies in degree $-n$ and equals to $Coker(d^{-n-1})$, so we obtained a finite resolution of $Coker(d^{-n-1})$ by finitely generated projective modules:

$$
\cdots \longrightarrow Z^{-n-2} \longrightarrow Z^{-n-1} \longrightarrow Z^{-n} \longrightarrow Coker(d^{-n-1}) \longrightarrow 0 .
$$

Therefore $Coker(d^{-n-1})$ is of finite projective dimension. $Z^{-n}$ and $Z^{-n-1}$ are finitely generated projective modules, so there are such (finitely generated projective) modules $P_0$ and $P_1$ and finitely generated free modules $F_0$ and $F_1$, that $Z^{-n} \oplus P_0 = F_0$ and $Z^{-n-1} \oplus P_0 \oplus P_1 = F_1$. So we have a resolution of $Coker(d^{-n-1})$ of the form

$$
\cdots \longrightarrow Z^{-n-2} \oplus P_1 \longrightarrow F_1 \longrightarrow F_0 \longrightarrow Coker(d^{-n-1}) \longrightarrow 0 .
$$

So $Coker(d^{-n-1})$ is finitely presented.

In fact, this looks like a standard exercise from a book on homological algebra, and it is usually more useful to do such an exercise by yourself than to post a question on MO.