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?. 

$x^6+3x^3+1-y^4=0$ is a quadratic equation.
Thus, there is an integer $n$ for which $$3^2-4(1-y^4)=n^2$$ or
$$n^2-4y^4=5$$ or
$$(n-2y^2)(n+2y^2)=5$$
and we have four cases only. 
