If $X\rightarrow Y\rightarrow Z$ follow a Markov chain, then we have the following properties$$I(X;Z)\leq I(X;Y)$$ where $I$ is the mutual information expression.
Intuitvely I agree. I want to formally prove it though.
So I try
$$I(X;Z)= H(X)-H(X|Z)$$ $$I(X;Y)= H(X)-H(X|Y)$$
Do we have that $$H(X|Z) >H(X|Y)$$ because of Markov chain?
Else I don't see how one can prove it.
Almost, but you need "greater than or equal to." We have:
$$ H(X|Y) = H(X|Y,Z) \leq H(X|Z) $$
where the first equality is from the Markov structure and the final inequality is because conditioning reduces entropy.
In more detail, to see how the Markov property works "backwards," notice that (assuming these point masses make sense):
\begin{align} Pr[X=x|Y=y,Z=z] &= \frac{Pr[X=x, Y=y, Z=z]}{Pr[Y=y,Z=z]} \\ &= \frac{Pr[X=x,Y=y]Pr[Z=z|X=x,Y=y]}{Pr[Y=y]Pr[Z=z|Y=y]} \\ &= \frac{Pr[X=x,Y=y]Pr[Z=z|Y=y]}{Pr[Y=y]Pr[Z=z|Y=y]} \: \: \: (*) \\ &= Pr[X=x|Y=y] \end{align}
where equation (*) uses the Markov property $Pr[Z=z|X=x,Y=y]=Pr[Z=z|Y=y]$.
If you are more comfortable using the Markov property forwards, you could also work out that $H(X|Y)=H(X|Y,Z)$ using $H(Z|X,Y)=H(Z|Y)$ (which is the "forward" Markov property when $X\rightarrow Y \rightarrow Z$) and some other basic properties of the entropy function, like $H(A,B) = H(A) + H(B|A)$.