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!

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...
