Sums of squares (Proof) [duplicate]
Currently Incomplete Answer
I am leaving it up for now to see if it helps anyone and so that if I think of something, I can edit it into the answer.
$$\begin{align} 2n^2+2n+1=m^3&\implies2n^2+2n=m^3-1\\ &\implies2n(n+1)=(m-1)(m^2+m+1)\\ &\implies m\equiv1\bmod4\\ &\implies2n(n+1)=4k((4k+1)^2+(4k+1)+1)\\ &\implies n(n+1)=32k^3+24k^2+6k\\ \end{align}$$ For this to use only integers, $n=2ak$ or $2ak-1$. Substituting into the equation yields $$2a^2k\pm a=a(2ak\pm1)=16k^2+12k+3$$