The Markov condition
$P(X_n=s|X_0=x_0,...,X_{n-1}=x_{n-1})=(X_n=s|X_{n-1}=x_{n-1})$ for all $n\geq1$ and all $s,x_0,...,x_{n-1} \in S$
is equivalent to:
$P(X_{n+m}=s|X_0=x_0,...,X_{n}=x_{n})=(X_{n+m}=s|X_{n}=x_{n})$ for any $n,m \geq1$
Is easy show the $\leftarrow$ implication $(m=1)$.
But, i'm stuck in the other way. If i try to use the theorem of total probability i get $P(X_{n+m}=s|X_{n+m-1}=x_{n+m-1})$. Can you help me?