It is best to index the rows and columns from $0$ to $2^m-1$, so $X_{ij}=i\,XOR\,j$.
We have a recursive construction for these matrices. Let $J$ denote the matrix of all ones. Let $X_m$ denote the matrix of size $2^m\times2^m$. Then we have in the block form $ X_{m+1}=\pmatrix{X_m&X_m+2^mJ\cr X_m+2^mJ&X_m\cr} $ for all natural numbers $m$.
On any row of $X_m$ all the integers in the range $[0,2^m-1]$ appear exactly once, so all the row sums are $\sum_{j=0}^{2^m-1}j=\frac12\,2^m(2^m-1)=2^{m-1}(2^m-1)$, and thus the vector $(1,\ldots,1)^T$ is an eigenvector belonging to this eigenvalue.
Assume that $v_m\in\mathbf{R}^{2^m}$ is an eigenvector of $X_m$ belonging to an eigenvalue $\lambda$. Furthermore, assume that either the entries of $v_m$ are all equal to one or that their sum is equal to zero. This latter requirement implies that $v_m$ is an eigenvector of the matrix $J$ belonging to the eigenvalue $\mu=2^m$ or to the eigenvalue $\mu=0$ depending which case applies.
Given this we can then construct two eigenvectors of $X_{m+1}$: $v_{m+1}^+=(v_m|v_m)$ and $v_{m+1}^-=(v_m|-v_m)$ by replicating the components of $v_m$ either without or with a sign change. We then see that $v_{m+1}^+$ is an eigenvector of $X_{m+1}$ belonging to the eigenvalue $\lambda+(\lambda+2^m\mu)=2\lambda+2^m\mu$, and $v_{m+1}^-$ is an eigenvector belonging to the eigenvalue $\lambda-(\lambda+2^m\mu)=-2^m\mu$.
Using this construction we can then recursively construct $2^{m+1}$ linearly independent eigenvectors of $X_{m+1}$ given $2^m$ linearly independent eigenvectors of $X_m$. This is because the component sum of $v_{m+1}^-$ is always equal to zero, and the component sum of $v_{m+1}^+$ is either zero (if that was the case with $v_m$) or it consists of all ones (ditto), so the assumption that these will be eigenvectors of $J$ will always hold. The starting point $m=0$ is covered by the all one vector, so basically this explains all the observations. Most of the time $\mu=0$, so we get the doubles of the eigenvalues of $X_{m+1}$ (with $v_{m+1}^+$) as well as zero (with $v_{m+1}^-$). The case where $\mu=2^m$ yields the positive eigenvalue (with $v_{m+1}^+$) as well as the negative eigenvalue with largest absolute value (with $v_{m+1}^-$).