TITLE: For what kind of positive integers $m+n+m^2+n^2 \mid m^3+n^3$ holds?
QUESTION [32 upvotes]: Determine all pairs of positive integers $(m, n)$ such that$$\frac{m^3+n^3}{m^2+n^2+m+n}$$is an integer.

This is a simpler version of a problem that I made a few years ago. There are infinitely many solutions. Such as parametric forms for $(m, n)$ that @Ivan Neretin has provided.
Let us consider $m>n$ due to symmetry (There is no solution for $m=n$). Let $m=ad$ and $n=bd$ for co-prime positive integers $a$ and $b$ and $a>b$. We get$$d^2(a+b)\big(a^2+b^2-ab\big)=k\big[d(a^2+b^2)+a+b\big]. \tag{1}$$
After @Geoff Robinson's first comment, I was looking for possible values of$$D=\gcd\big(m+n, m^2+n^2+m+n\big),$$which is$$D=\gcd\big(m+n, (m+n)^2-2mn+m+n\big)=\gcd(m+n, 2mn).$$and$$D=\gcd\big(d(a+b), 2d^2ab)=d \gcd(a+b, 2d).$$ Since $\gcd(a+b, ab)=1$. 
Then I used equation $(1)$ and the relation between $d$ and $\gcd(a+b, 2d)$, for finding a relation between $d$ and $a+b$. I think it can be the key for solving the problem.
After looking at patterns using a code, I realized that actually there is a relation between $d$ and $a+b$. It is interesting that for most of the solutions we get $\gcd(a+b, 2d)=2d$ and $\frac{d}{a+b}=\frac{1}{4}$.

REPLY [2 votes]: For the equation.
$$x^3+y^3=q(x^2+y^2+x+y)$$
We use the solutions of the equation Pell.
$$p^2-(3k-2t)(2t-k)s^2=1$$
Then the solution can write.
$$x=((3k-2t)s+p)2ts$$
$$y=\frac{t}{k-t}(p^2+2(2k-t)ps+(3k-2t)ks^2)$$
$$q=\frac{t^2}{k(k-t)}(p^2+2(3k-2t)ps+(3k-2t)^2s^2)$$
And again.
$$y=((3k-2t)s-p)2ts$$
$$q=\frac{4(3k-2t)}{k}t^2s^2$$
All the numbers can have any sign.