As functions of $k$, it appears that both sides satisfy the recurrence
\begin{align}
& 4(-n+2\,k+1) (-n+2k) A(n,k) \\[6pt]
& {} + (8k^2-8\,kn+n^2+10k-9n) A(n,k+1)\\[6pt]
& {} + (k+2)(-n+k) A(n,k+2)=0
\end{align}
with $A(n,0) = 1$, $A(n,1) = n-2$.