Integer solutions of $m^2+(m+1)^2=n^4+(n+1)^4$
$$n^4+(n+1)^4=2n^4+4n^3+6n^2+4n+1$$ $$=2(n^4+2n^3+3n^2+2n+1)-1=2y^2-1$$ where $y=n^2+n+1$, so you want $2y^2-1$ to be a sum of consecutive squares. It's not going to happen (except for the instances you have already found): $(y-1)^2+y^2$ is too small, and $y^2+(y+1)^2$ is too big.
Solving $$m^2+(m+1)^2=n^4+(n+1)^4$$ for $m$ gives you $$m=\frac{-1\pm\sqrt{4n^4+8n^3+12n^2+8n+1}}{2}$$ However, for $n\not=0,-1$, since $$2n^2+2n+1\lt \sqrt{4n^4+8n^3+12n^2+8n+1}\lt 2n^2+2n+2,$$ we know that $\sqrt{4n^4+8n^3+12n^2+8n+1}$ cannot be an integer.
Hence, we have $n=0,-1$.