It is apparently assumed that the $Z_{ij}$'s are independent, as we will do here -- since otherwise hardly anything can be said. Suppose also that $m\ge2$ and $0<p<1$. 

The $ab$-entry of the matrix $Y:=XX^\top$ is 
\begin{equation}
	Y_{ab}=\sum_{r\in[m]}X_{ar}X_{br}
	=\sum_{r\in[m]}\frac{Z_{ar}Z_{br}}{\sum_{k\in[m]}Z_{ak}\sum_{k\in[m]}Z_{bl}},
\end{equation}
where $[m]:=\{1,\dots,m\}$ and $a,b$ are in $[n]$. 
So, 
\begin{equation}
	EY_{ab}=m \,E\frac{Z_{a1}Z_{b1}}{\sum_{k\in[m]}Z_{ak}\sum_{k\in[m]}Z_{bl}}.
\end{equation}
If $a\ne b$ then 
\begin{equation}
\begin{aligned}
EY_{ab}&=m\,E\frac{Z_{a1}}{\sum_{k\in[m]}Z_{ak}} \, E\frac{Z_{b1}}{\sum_{k\in[m]}Z_{bl}} \\ 
&=m \Big(E\frac{Z_{a1}}{\sum_{k\in[m]}Z_{ak}}\Big)^2 \\ 
&=m p^2\Big(E\frac1{1+\sum_{k\in[m]\setminus\{1\}}Z_{ak}}\Big)^2 \\ 
&=m p^2\Big(E\frac1{1+B_{m-1,p}}\Big)^2 \\ 
&=\frac{\left(1-q^m\right)^2}{m}, 
\end{aligned}
\end{equation}
where $q:=1-p$ and $B_{m-1,p}$ is a binomial random variable with parameters $m-1,p$. 
Similarly, the diagonal entries of the matrix $EY$ are 
\begin{equation}
\begin{aligned}
EY_{aa}&=m E\Big(\frac{Z_{a1}}{\sum_{k\in[m]}Z_{ak}}\Big)^2 \\ 
&=m p\,E\Big(\frac1{1+\sum_{k\in[m]\setminus\{1\}}Z_{ak}}\Big)^2 \\ 
&=m p\,E\Big(\frac1{1+B_{m-1,p}}\Big)^2 \\ 
&=mp q^{m-1} \, _3F_2\left(1,1,1-m;2,2;-\frac{p}{q}\right), 
\end{aligned}
\end{equation} 
where $_3F_2$ is the hypergeometric function.