<s>[INST] Diophantine equation $x^2 + xy + y^2 = \left({{x+y}\over{3}} + 1\right)^3$. Solve in integers the equation$$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3.$$
 [/INST] Hint $1$ :
Let $t=\frac{x+y}{3}\in \mathbb Z$ the equation becomes:
$$y^2-3ty+(3t)^2-(t+1)^3=0 $$
a quadratic equation on $y$ which is soluble up to the condition $4t+1$ is a square.
Hint $2$: $\Delta_y=(t-2)^2(4t+1)$
Solutions $(t,y)=(a^2+a,-a^3+3a+1),(a^2+a,a^3+3a^2-1)$ for any parameter $a$
And it's your turn to do some work!

Edit
Because my answer is down-voted and I don't see the reasons, I will write a full solution based on my little hints below:

Claim If $(x,y)$ are integer solutions to the equation  :
  $$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3\tag 1$$ then
  $(x,y)\text{ or } (y,x) \in\left\{\left(-a^3+3a+1,a^3+3a^2-1\right)\big | a\in \Bbb Z \right\}$

Proof
Let $(x,y)$ be a solution to the given equation, and let $t=\frac{x+y}{3}\in \Bbb Z$ so that:
$$y^2-3ty+(3t)^2-(t+1)^3=0 \tag 2$$
and this is a quadratic equation on $y$ so let's compute its discriminant:
$$\Delta_y=(3t)^2-4\left((3t)^2-(t+1)^3\right)=4(t+1)^3-27t^2=(t-2)^2(4t+1)\tag 3$$
so this equation has an integer solution $y$ only if $\Delta_y$ is a square of an integer, and this is equivalent to $4t+1$ is a square of an integer which implies that $t=a^2+a$ for some integer $a$. and because the equation $(2)$ is quadratic and we know the value of its discriminant we can find the two solutions on $y$:
$$\begin{align}y&=&\frac12\left(3t-(t-2)\sqrt{4t+1}\right)&=-a^3+3a+1\tag 4\\ y&=&\frac12\left(3t+(t-2)\sqrt{4t+1}\right)&=a^3+3a^2-1 \tag 5\end{align}$$
now that we found the value of $t$ and $y$ we can find the value of $x=3t-y$ which gives:
$$\left\{\begin{matrix}
 x=a^3+3a^2-1&\text{ and }& y=-a^3+3a+1 &\text{ or }\tag 6\\ 
 x=-a^3+3a+1&\text{ and } & y=a^3+3a^2-1 
\end{matrix}\right.$$
Finally:
$$(x,y)\text{ or } (y,x) \in\left\{\left(-a^3+3a+1,a^3+3a^2-1\right)\big | a\in \Bbb Z \right\}\tag 7$$
</s>