There are infinitely many integer solutions of equation $(1).$  

$$y^2=\frac{n(n+1)(n+2)}{6}+\frac{m(m+1)(m+2)}{6}\tag{1}$$
Substitute $n = s - m$ to equation $(1)$ then we get  
$$6y^2 = (3s+6)m^2+(-3s^2-6s)m+s^3+2s+3s^2$$  

Let $s = 2t$ and $x = m - t$ then we get  

$$3y^2 = (3t+3)x^2+t(t+2)(t+1)$$  

Let $t = u^2-1$ then we get  

$$y^2-u^2x^2 = \frac{(u^2-1)(u^2+1)u^2}{3}$$  

Hence we get $(x,y)= \left(\Large{\frac{u^2+2}{3},\frac{2u^3+u}{3}}\right)$  

To make $x$ and $y$ integers, let $u = 3k + 1$ then we get a parametric solution of euation $(1).$  

$(m,n,y)=(\ (6k+1)(2k+1),\ 6k^2+4k-1,\ (3k+1)(6k^2+4k+1) \,)$  
$k$ is arbitrary integer. 

Similarly, $u = 3k + 2$ then we get another solution.  

$(m,n,y)=(\ (2k+1)(6k+5),\ 6k^2+8k+1,\ (3k+2)(6k^2+8k+3) \,)$  

Numerical example with $k\leqq 10.$  

$(m,n,y)=(21, 9, 44),(65, 31, 231),(133, 65, 670),(225, 111, 1469),(341, 169, 2736),(481, 239, 4579),(645, 321, 7106),(833, 415, 10425),(1045, 521, 14644),(1281, 639, 19871),$    
$(5, 1, 6), (33, 15, 85), (85, 41, 344), (161, 79, 891), (261, 129, 1834), (385, 191, 3281), (533, 265, 5340), (705, 351, 8119), (901, 449, 11726), (1121, 559, 16269), (1365, 681, 21856)$