Solving $m^3-n^3=2mn+8$ We have to find integer solutions to the given equation, this is what i tried :-
For ease, let us denote $x=-m, \enspace n=-y$, and then we are basically considering $x,m,y,n$ as nonnegative wherever they occur below :- 
We have four cases :-
CASE $1$ : $mn<0$ and $x^3+n^3=-(8+2xn)$
This case is clearly rejected as $x,n$ are both positive.
CASE $2$ : $mn<0$ and $m^3+y^3=8+2my$
For this we apply AM - GM to get,
$$9 + 2my = m^3 + y^3 + 1 \geq 3my \implies my \le 9 $$
Case bash (sadly) gives that the solutions to this case are
$$(m,y)=(0,2), \enspace (2,0), \enspace (2,2)$$

Now we are left with two cases
CASE $3$ : $mn>0$ and $m^3-n^3=8-2mn$
CASE $4$ : $mn>0$ and $x^3-y^3=2xy-8$
Now how to finish off these last $2$ cases?
Thanks in advance!
 A: Observe that $x^3 - (x-1)^3 > 2x(x-1) + 8$ for all $x$ except $-2 ≤ x ≤ 3$ and that $x^3 - x^3 < 2x(x)+ 8$ for all $x$. From this, the equation lies between $n = m-1$ and $n = m$, hence the only integer solutions are when $-2 ≤ m ≤ 3$.
A: Another way to solve this is to introduce $x=m-n$. Then you have quadratic equation on $n$:
$$(3x-2)n^2+(3x^2-2x)n +x^3-8=0$$
This equation has integer solution if it discriminant is perfect square:
$$(3x^2-2)^2-4(3x-2)(x^3-8) = d^2$$
In particular $$3x^4-8x^3+12x^2-96x+60\leq 0$$ which can not occurre many times...
A: With a little help from ultralegend5385's hint.
Let $m, n \in \mathbb{Z}$. Factoring the LHS, we obtain  $$m^3-n^3 = (m-n)(m^2+mn+n^2)$$
Then our equation becomes
$$ m^3-n^3=(m-n)(m^2+mn+n^2)=2mn+8=2(mn+4)$$
$$ (m-n)(m^2+mn+n^2)=2(mn+4)$$
We can split this equation into two cases:
Case 1: $$(m-n) = 2 \tag{1}\label{eq1}$$
$$(m^2+mn+n^2)=(mn+4)\tag{2}\label{eq2}$$
Equation \eqref{eq1} becomes $m=n+2$, then substituting into equation \eqref{eq2}, we get
$$ (n+2)^2 + (n+2)n + n^2 = (n+2)n+4$$
$$ (n^2 + 4n + 4) + n^2+2n + n^2 = n^2 + 2n+ 4$$
$$2n^2+4n+4=4$$
$$n^2+2n=0$$
$$n(n+2)=0$$
So then $n=0, -2$ satisfy the equation. Then $n=0\implies m=2$ and $n=-2 \implies m=0$. Two solutions to our original equation are $(m,n)=(2,0)$ and $(m,n)=(0, -2)$.
Case 2: $$(m-n) = (mn+4) \tag{3}\label{eq3}$$
$$(m^2+mn+n^2)= 2\tag{4}\label{eq4}$$
From Equation \eqref{eq3}, we get that $(m-n)$ must be even. If m and n are of opposite parity, their product must be even so the RH side will be even but the LH side will be odd which is impossible. If they are both odd then their product is odd so the RH side will be odd but their difference will be even again impossible. Therefore m and n must both be even.
Let $k\in \mathbb{Z}$
$$ 2k=mn+4$$
$$ 2k-4=mn$$
$$2(k-2)=mn \implies$$ $$m=2 \text{ and } n=(k-2) \tag{5}\label{eq5}$$ or
$$m=(k-2) \text{ and } n=2 \tag{6}\label{eq6}$$
Let's substitute equations \eqref{eq5} into equations \eqref{eq4}, so that we can attempt to solve for k
$$ (2^2 + 2(k-2)+(k-2)^2)=2$$
$$ 4 + 2k - 4 + (k^2 - 4k + 4) = 2$$
$$ k^2-2k+2=0\tag{7}\label{eq7}$$
Equation \eqref{eq7} is in the form of a quadratic equation, so we can apply the quadratic formula.
When we do that we get imaginary roots, so there is no solution for k in the integers.
We can observe that equation \eqref{eq4} is symmetric in terms of m and n, so equations \eqref{eq6} yield the same result.
Therefore, our only solutions for the equation $m^3-n^3=2mn+8$ is
$(m,n)=(2,0)$ and
$(m,n)=(0, -2)$.
P.S. Sorry for the formatting this is my first longer post
A: The trivial solutions from Wolfram Alpha
are:
$$m^3-n^3=2mn+8\quad \implies \quad (m.n)\in\big\{(0,-2),\space (2,0)\big\}$$
We can try to factor and cancel to no avail.
$$m^3-n^3=(m-n)(m^2+mn+n)=2mn+8$$
Brute force shows no other solutions for
$\quad-10^5\le m,n \le 10^5\quad$ (the limit of $15$-figits)
and by inspection we can see that these are likely the only solutions where
$$0-(-2)^3=2(0)(-2)=8\qquad\text{ or }
\qquad 2^3-0=2(2)(0)+8$$
