All sums below are by non-negative $i,j$ unless opposite is stated explicitly. 

Denote $h(x,y)=\sum \binom{i+j}i x^iy^j=\frac1{1-(x+y)}$, $f(x,y)=\sum \binom{i+j}i^2 x^iy^j$. We want to prove that $2xyf^2(x^2,y^2)$ is an odd (in both $x$ and $y$) part of the function $h(x,y)$. In other words, we want to prove that $$2xyf^2(x^2,y^2)=\frac14\left(h(x,y)+h(-x,-y)-h(x,-y)-h(-x,y)\right)=\frac{2xy}{1-2(x^2+y^2)+(x^2-y^2)^2}.$$
So, our identity rewrites as $$f(x,y)=(1-2(x+y)+(x-y)^2)^{-1/2}=:f_0(x,y)$$
This is true for $x=0$, both parts become equal to $1/(1-y)$. Next, we find a differential equation in $x$ satisfied by the function $f_0$. It is not a big deal: $$\left(f_0(1-2(x+y)+(x-y)^2)\right)'_x=(x-y-1)f_0.$$
Since the initial value $f_0(0,y)$ and this relation uniquely determine the function $f_0$, it remains to check that this holds for $f(x,y)$, which is a straightforward identity with several binomials. Namely, comparing the coefficients of $x^{i-1}y^j$ we get
$$ i\left(\binom{i+j}j^2-2\binom{i+j-1}j^2-2\binom{i+j-1}i^2+\binom{i+j-2}i^2+\binom{i+j-2}j^2-2\binom{i+j-2}{i-1}^2\right) 
$$ 
for $(f(1-2(x+y)+(x-y)^2))'_x$ and $$\binom{i+j-2}j^2-\binom{i+j-1}j^2-\binom{i+j-2}{j-1}^2$$
for $(x-y-1)f$. Both guys are equal to $$-2\frac{j}{i+j-1}\binom{i+j-1}{j}^2.$$