Solve the Diophantine equation $x^6 + 3x^3 + 1 = y^4$

$x^6+3x^3+1-y^4=0$ 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.

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:

$$(t^3-2)^2<y^4<(t^3-1)^2$$