Prove by induction on $m$ that for all $n,m\in\N_0$: $\E^x(f(X_{n+m})|\F_n)=P^mf(X_n)$.
Assume $\E^x(f(X_{n+m})|\F_n)=P^mf(X_n)$ for all $n\in\N_0$, then by the Markov property:
$$
\E^x(f(X_{n+m+1})|\F_n)
=\E^x(\E^x(f(X_{n+m+1})|\F_{n+1})|\F_n)
=\E^x(P^mf(X_{n+1})|\F_n)
=PP^mf(X_n)=P^{m+1}f(X_n)~.
$$