Solution of Diophantine Equation $1+a^6=x^2$

Question

Does the equation $1+a^6=x^2$ have any other integer solutions except the trivial one $a=0,x=1$?

Answer 1

It's $$(x-a^3)(x+a^3)=1$$ and solve two systems: $$x-a^3=1$$ and $$x+a^3=1$$ or $$x-a^3=-1$$ and $$x+a^3=-1,$$ which gives the answer for $(x,a)$: $$\{(1,0), (-1,0)\}.$$

Answer 2

$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $n\ge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).