Solving $y^3=x^3+8x^2-6x+8$

Since we have$$y^3-x^3=2(4x^2-3x+4)$$ there exists an integer $k$ such that $$y-x=2k\iff y=x+2k.$$ So, we have $$(x+2k)^3-x^3=2(4x^2-3x+4)\iff (3 k-4) x^2+(6 k^2+3) x+4k^3-4=0\tag1$$ Now we have $$D=(6k^2+3)^2-4(3k-4)(4k^3-4)\ge 0\iff -12 k^4+64 k^3+36 k^2+48 k-55\ge 0$$$$\iff 12 k^4-64 k^3-36 k^2-48 k+55\le 0$$

Here, let $f(k)=12 k^4-64 k^3-36 k^2-48 k+55=4k[k\{k(3k-16)-9\}-12]$.

Now, for $k\ge 6$, we have $$3k-16\ge 2\Rightarrow k(3k-16)\ge 2k\ge 12$$$$\Rightarrow k\{k(3k-16)-9\}-12\ge 3k-12\gt 0\Rightarrow f(k)\gt 0.$$

Also, for $k\le -1$, we have $$f(k)=12 k^4-36k^2(k+1) -48 k+55-28k^3\gt 0.$$

Hence, we have $k=0,1,2,3,4,5.$

Then, from $(1)$, you can find integer roots $x$ for each $k$.

Using Gerry Myerson's clever inequality observation, we have two possibilities.

Case 1: $y=x+1$. Then by substitution and simplification, we must solve $$5x^2-9x+7=0,$$ which has no integer solutions, as seen by using the quadratic equation.

Case 2: $y=x+2$. This yields the [factored] quadratic $$2x(x-9)=0.$$ Evidently, the only two integer solutions are $x=0$ and $x=9$, yielding the solutions $(x,y)=(0,2)$ and $(x,y)=(9,11)$.

There are no other solutions.

We have

$y^3 − (x + 1)^3 = x^3 + 8x^2 − 6x + 8 − (x^3 + 3x^2 + 3x + 1) = 5x^2 − 9x + 7$.

Consider the quadratic equation

$5x^2 − 9x + 7 = 0$.

The discriminant of this equation is

$D = 92 − 4 × 5 × 7 = −59 < 0$ and hence the expression $5x^2 − 9x + 7$ is positive for all real values of x.

We conclude that $(x + 1)^3 < y^3$ and hence $x + 1 < y$.

On the other hand we have

$(x + 3)^3 − y^3 = x^3 + 9x^2 + 27x + 27 − (x^3 + 8x^2 − 6x + 8) = x^2 + 33x + 19 > 0$ for all positive $x$.

We conclude that $y < x + 3$.

Thus we must have $y = x + 2$. Putting this value of y, we get $0 = y^3 − (x + 2)^3 = x^3 + 8x^2 − 6x + 8 − (x^3 + 6x^2 + 12x + 8) = 2x^2 − 18x$.

We conclude that $x = 0$ and $y = 2$ or $x = 9$ and $y = 11$