Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$. 
Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.

What I Tried:- Given this expression I had no idea to claim something out of it, because it looks like we need to $437$ in terms of sums of $2,3$ (the $6$ does not matter much here). I could break $(8n^3 + m^3)$, but it didn't help much.
Instead I simplified the whole expression and got :-
$$2m^3 + 6m^2 + 6m + 16n^3 - 36n^2 + 27n = 437.$$
$$\rightarrow 2m^3 + 6m^2 + 6m + 16n^3 - 36n^2 = 437 - 27n.$$
From here, $2 \vert (437 - 27n)$, so $n$ is odd. Let $n = (2k + 1)$. Substitute this back to the equation to get :-
$$64k^3 + 24k^2 + 3k + m^3 + 3m^2 + 3m = 215.$$
Now I am stuck here, it is given $(k,m)$ are integers, so they can range to negative values also. So I was not able to make an upper bound on $k$, and I have no other idea what I can claim from here.
Can anyone help me? Thank You.

First, substitute $x=m+1$ and move the constant term to the left side:
$$2x^3+16n^3-36n^2+27n=439$$
Then, note that $n$ must be odd.  We can replace $n=2y+1$ and divide out by $2$ and get:
$$x^3+64y^3+24y^2+3y=216$$
Some algebraic manipulation (multiplying both sides by $8$, adding $1$ to both sides, factoring, and then replacing $z=8y+1$):
$$8x^3+512y^3+192y^2+24y=1728$$
$$8x^3+512y^3+192y^2+24y+1=1729$$
$$(2x)^3+(8y+1)^3=1729$$
$$(2x)^3+z^3=1729$$
Expressing $m$ and $n$ in terms of $x$ and $z$ gives $m=x-1$, $n=\frac{z-1}{4}+1$.
$1729$ is famously the Ramanujan-Hardy taxicab number, the smallest natural number that can be written as the sum of two cubes in two different ways.  $1729=1^3+12^3=10^3+9^3$.  This results in possible $(x,z)$ of $(6,1)$ or $(5,9)$.  In terms of $(m,n)$, we have $(4,3)$ and $(5,1)$.
Therefore, our possible values for $mn$ are $5$ and $12$.
