can I finish this solution by this method, if I can please tell me how $m^2+m=n^3+n^2+n$, find solutions in natural numbers. so this was the problem and answer is that it doesn't have solutions in natural numbers.
It was pretty hard to think about this problem, I wanted to analyze the ends
of the numbers, for example, if $m$ ends on 2, then $m^2$ ends on 4, so their sum ends on 6.
First of all, it is easy to prove that n is even. since the left side is even, then right side must be the even as well, so $n$ must be even.  So I analyzed when n is even, at what number sum n^3+n^2+n ends on, for example if $n$ ends on 2 then $n^2$ ends on 4.
Then I analyzed when $n$ ends on 4,6,8. sum ends on 4 or 8. Then I analyzed ending of $m$ and $m^2$ then I got that $m^2+m$ doesn't end on 4 or on 8, it ends on 0 in 3 cases, when m ends on 4,5,9  $m+m^2$ ends on 0. In other cases $m$ doesn't end on 4 or on 8 so I supposed that $m+m^2$ doesn't equal $n+n^2+n^3$, for natural numbers, but unfortunately and I am very sad right now, I forgot to analyze when $n$ ends on 0. when $n$ ends on 0, it sum $n+n^2+n^3$ ends on 0 and unfortunately, I have to analyze 3 cases for $m$, when sum $m+m^2$ ends on 0.
So I really need your opinion about this can I finish this problem after analyzing ending 0 situations, please answer me it will be very helpful for me, I really want to defend this solution.
 A: Another approach:
$$m^2+m=\frac{m^3-1}{m-1}-1$$
$$n^3+n^2+n=\frac{n^4-1}{n-1}-1$$
Therefore:
$$\frac{m^3-1}{m-1}=\frac{n^4-1}{n-1}$$
Due to Fermat's little theorem:
$n^4-1 \equiv 0 \mod (5)$,$\rightarrow n^4-1=5t$
$m^3-1$ can be a multiple of 5 with some condition:
$m\equiv (0, 1, 2, 3, 4)\ mod (5)$
If $m \equiv 1 \ mod (5)$, then:
$m^3-1 \equiv 0 \ mod (5)$
But not all multiples of 5 are of the form $n^4-1$ unless $m=a^4$ such that $(a^4)^3-1= (a^3)^4-1$. I n this case $n=a^3$.But this does not satisfy the equation, for example let $a=2$ we have:
$m=2^4=16$, $n=2^3=8$
$m^3-1=2^{12}-1= 8^4-1$
But:
$m^2+m=16^2+16=272$
$n^3+n^2+n=8^3+8^2+8=584$
That is this equation can not have integer solution.
A: I am not sure whether your solution can be defended or not, since we could have both $m$ and $n$ to be divisible by $10$, and if you try to remove common factors, the argument won't hold anymore.
One way to go about solving this problem might be to try to find a nice factorization instead. We have:
$$m^2+m-n^2-n=n^3 \implies (m-n)(m+n+1)=n^3$$
Now, writing $m-n=a$ gives $a(a+2n+1)=n^3$. If there exists a common prime factor for $a$ and $a+2n+1$, it would divide $2n+1$, and thus cannot divide $n$, which would be a contradiction. Thus, we must have $a$ and $a+2n+1$ to be relatively prime, and since their product is a cube, they are both perfect cubes. Writing $a=x^3$ gives:
$$x^3(x^2+2n+1)=n^3$$
Since $x^3 \mid n^3$, we can write $n=xk$. This gives:
$$x^3+2xk+1=k^3$$
Again, writing $k=x+y$ gives:
$$x^3+2x^2+2xy+1=x^3+3x^2y+3xy^2+y^3$$
I believe you can continue from here...
