I don't know a reference. One way to show the eigenvalues starts from the observation (which can be proved using generating functions) that $\sum_{i=0}^n {i\choose k} A_{i,j}={2n+1 \choose n-k} {j+k \choose k}={2n+1 \choose n-k}\,\sum_{\ell=0}^k {k\choose \ell} { j \choose \ell}$ (where $A$ is the matrix above). With the row vectors $\mathbf{v}_k$ with coordinates $\mathbf{v}_k(i)={i \choose k}$ that is $$\mathbf{v}_k A={2n+1 \choose n-k}\left(\sum_{\ell=0}^k {k \choose \ell} \mathbf{v}_\ell\right).$$ The rest is routine.

I don't know a reference. One way to show the eigenvalues starts from the observation (which can be proved using generating functions) that $\sum_{i=0}^n {i\choose k} A_{i,j}={2n+1 \choose n-k} {j+k \choose k}={2n+1 \choose n-k}\,\sum_{\ell=0}^k {k\choose \ell} { j \choose \ell}$ (where $A$ is the matrix above). With the row vectors $\mathbf{v}_k$ with coordinates $\mathbf{v}_k(i)={i \choose k}$ that is $$\mathbf{v}_k A={2n+1 \choose n-k}\left(\sum_{\ell=0}^k {k \choose \ell} \mathbf{v}_\ell\right).$$ The rest is routine.

ADDED: (for the record)
With some patience one finally finds that $$\mathbf{e}_k=\sum_{j=0}^k (-1)^j{n-j \choose k-j}{k+j \choose j}\mathbf{v}_j$$ is an eigenvector to the eigenvalue ${2n+1 \choose n-k}$.