m is odd since if n is even, even+even+odd and if n is odd, odd+odd+odd. Now let's look at this mod 3. Cubing the term on the right yields 0 if you put in 0, 1 if you put in 1, and 2 if you put in 2. The left hand side will have 0+0+7 or equaling to 1 mod 3. In other words, 3x+1 are the form our numbers come in. We substitute 3x+1 for m where x is even. (3x+1)^3=27x^3+27x^2+9x+1=3n^2+3n+7. We simplify to get 9x^3+9x^2+3x-2=n^2+n. Now let's consider this in mod 3 again. For the terms on the right, we get 0+0+0-2 or 1 mod 3. The left hand side is 0 when 0 is inserted, 2 when 1 is inserted, and 0 where 2 is inserted. In other words, it never equals one mod 3 so substituting 3x+1 for m causes an answer to be nonexistant but that's the only form the answers come in from the first part of the proof. In other words, there's a contradiction so there is no positive integers m and n.

Here's why I don't think it's right. If you do the mod 3 thing immediately after we substitute 3x+1 in, we don't have a problem. But when we analyze with mods after simplification, we get the contrdiction.

mod 9.

We can change it into $(n+1)^3-n^3+6=m^3$ Notice that the residue of a cube mod 9 can only be $\pm 1,0$ . And for difference of consecutive cube , the only possiblity is $(n+1)^3-n^3\equiv 1,-2\mod 9$ . So LHS$\equiv 4,7\mod 9$ which obviously shows no solution for $m^3$ .