Find all the integral solutions to $x^6-y^6+3x^4y-3y^4x+y^3+3x^2+3x+1=0$

Given: $x^6-y^6+3x^4y-3y^4x+y^3+3x^2+3x+1=0$

$\implies(x+1)^3+(x^2+y)^3=(x+y^2)^3$

By Fermat's Last Theorem, since $x,y\in\mathbb{Z}^+$ , this equation has no solution.

Hint:

You can rearrange this as $$(x+1)^3-x^3=y^6-x^6+3xy(y^3-x^3)=(y^3-x^3)(y^3+3xy+x^3)$$

And note that $x,y\in \mathbb Z^+$

I think it can be done directly, without reducing to FLT. A bit of algrebra shows that $|x^6 - y^6|$ dominates the lower order terms for all except a very small finite collection of $x$ and $y$ (which can be checked by hand):

First, it's easy to see that any solution must have $y \gt x \gt 0$, so write $y = x + a$, with $a \gt 0$ also.

Now substitute and expand everything. I get, $$ y^6 - x^6 = 6ax^5 + 15a^2x^4 + 20a^3x^3 + 15 a^4x^2 + 6a^5x + a^6 $$ while, if my pen-and-paper calculation is correct, $$ 3x^4y - 3xy^4 + 3x^2 + 3x + 1 = -(3a^4-1)x - (12 a^3 - 1) x^2 - 18 a^2 x^3 - 9 a x^4 + 1 $$ Comparing terms in the two expressions above, we see that the sixth order terms are strictly larger (in absolute value) than lower order term involving the same power of $a$, (except for very small $a,x$). E.g., $$ 18 a^2 x^3 < 15 a^2 x^4 \quad \textrm{unless} \quad a,x < 2 $$ $$ |(3a^4 - 1)x| < 15 a^4 x^2 \quad \textrm{if} \quad a,x \ge 1 $$ and also, $a^6 > 1$ for $a > 1$.

All the sixth order terms have the same sign. Then we have that the sum of the sixth order terms is strictly larger than the absolute sum of the lower order terms. The remaining cases (I think just $a,x \in {1,2}$) can be checked by hand.