Solve the Diophantine equation $x^6 + 3x^3 + 1 = y^4$ 
Find all pairs $(x, y)$ of integers, such that $x^6 + 3x^3 + 1 = y^4$. 

My solution:
Claim: The pair $(0, 1)$ are the only solutions.
Proof. Suppose there exists other solutions for $y \gt 1$ and $x \gt 0$, then I shall show that such pairs are impossible if $x$ and $y$ are integers. Let us Factorize the given equation as follows. 
$x^6 + 3x^3 = y^4 - 1$
$x^6 + 3x^3  = (y^2)^2  - 1^2$
$x^6 + 3x^3 =(y^2 - 1)(y + 1)(y^2 + 1)$
$x^3(x^3+ 3) = (y - 1)(y + 1)(y^2 + 1) . . . (*)$
Because  the numbers $(y - 1), (y + 1)$, and $(y^2 + 1)$ are all distinct, it follows that their products can never be a cube, hence, from $(*)$, we obtain the system.
$(y - 1)(y + 1)(y^2 + 1) = x^3 + 3$, and
$x^3 = 1$,
Which is equivalent to:
$(y - 1)(y + 1)(y^2 +) = 4 . . .(**)$
Since $(y - 1) \gt 0$, then $y$ is minimum if and only if $y = 2$ which clearly does not satisfy $(**)$. Hence we must have $x = 0$ and $y = 1$. 
Please, Is there any mistake in my solution?. 

Case 1. If $x= 0$ we have $y^4=1$ thus $y=\pm 1$
Case 2. If $x>0$ then
$(x^3+1)^2= x^6+2x^3+1<  x^6 + 3x^3 + 1 = y^4$ 
$ y^4 = x^6 + 3x^3 + 1< x^6+4x^3+4 = (x^3+2)^2$ 
So we have:$$(x^3+1)^2< y^4 <(x^3+2)^2$$ 
So $x^3+1< y^2 < x^3+2$ a contradiction. 
Case 3. If $x<0$ then write $x=-t$ and $t>0$. Now we have to sove:
$$t^6 - 3t^3 + 1 = y^4$$ and this can be done in similar fashion:
$$(t^3-2)^2<y^4<(t^3-1)^2$$
