Find all positive integers x such that $x^4-8x+16$ is a full square

Question

Find all positive integers x such that $x^4-8x+16$ is a full square.

I presented it as $(x-4)^2+x^4-x^2$ and found that $x = +1$ satisfy the condition. Then I equalized it $(x-4)^2+x^4-x^4=a^2+2ab+b^2$ then $a=x-4$ and nothing else happened. Help prove that there are no more solutions or find the rest of the solutions

Answer 1

For $x \ge 3$, we have:

$$x^4-8x+16=x^4-8(x-2)<x^4$$

and

$$x^4-8x+16-(x^2-1)^2=2(x-2)^2+7 > 0$$

Therefore $x^4>x^4-8x+16>(x^2-1)^2$ for $x\geq 3$ (it is bounded by two consecutive perfect squares).

It remains to check $x=1$ and $x=2$, which are indeed solutions.

Answer 2

Hint:

For $x>2$, $(x^2-1)^2<x^4-8x+16<(x^2)^2$.