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Michael Hardy
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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 $A(n,0) = 1$, $A(n,1) = n-2$.

Robert Israel
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