Bijective proof for $\sum\limits_{\substack{n_1+n_2=n\\m_1+m_2=m}}\binom{n_1+m_1}{n_1}^2\binom{n_2+m_2}{n_2}^2=\frac{1}{2}\binom{2n+2m+2}{2n+1}$?

Question

Let $G(x, y)$ be the bivariate generating function for $\binom{i+j}{i}^2$, that is $$ G(x, y) = \sum\limits_{i,j} \binom{i+j}{i}^2 x^i y^j. $$ It can be shown that $$ G(x^2, y^2) = \frac{1}{\sqrt{(1-(x-y)^2)(1-(x+y)^2)}}. $$ Raising both parts to the power $2$ and regrouping leads to the identity $$ \sum\limits_{\substack{n_1+n_2=n\\m_1+m_2=m}}\binom{n_1+m_1}{n_1}^2\binom{n_2+m_2}{n_2}^2=\frac{1}{2}\binom{2n+2m+2}{2n+1}. $$ The left-hand side seems to enumerate pairs of paths from $(0, 0)$ to $(n, m)$ that share a specific point $(n_1, m_1)$. Is there a bijective proof that this number is counted by the right-hand side? Is this a well-known identity?