How do I prove 8(x)! + (2x-1)² where x is an integer greater than or equal to 3, is never a perfect square.

Question

$8(x)! + (2x-1)^2 = a^2$, so $8(x)! = a^2-(2x-1)^2$, for $x \geq 8$, $64|\text{LHS}$, but we can see 8 doesn't divide both of $(a-2x+1)(a+2x-1)$, so 64|one of them, so both one of them is $0 \pmod {64}$ and thus $0 \pmod{8}$, so $(a+1)$ and $(a-1)= 0\pmod8$, so $a= \pm1 \pmod8$, please solve it or hint further:)