Let $ r $ and $ s $ be the $ x $-coordinates of the two points of tangency. As such, they will be double roots of the polynomial \[(x^4 + ax^3 + x^2 + bx + 1) - x^2 = x^4 + ax^3 + bx + 1.\]Hence, \begin{align*} x^4 + ax^3 + bx + 1 &= (x - r)^2 (x - s)^2 \\ &= (x^2 - 2rx + r^2)(x^2 - 2sx + s^2) \\ &= x^4 - (2r + 2s) x^3 + (r^2 + 4rs + s^2) x^2 - (2r^2 s + 2rs^2) x + r^2 s^2.\end{align*}Matching coefficients, we get \begin{align*} r^2 + 4rs + s^2 &= 0, \\ r^2 s^2 &= 1.\end{align*}From $ r^2 s^2 = 1, $ either $ rs = 1 $ or $ rs = -1 $. But $ 4rs = -(r^2 + s^2) $ is nonpositive, so $ rs = -1 $. Then \[r^2 + s^2 = 4.\]Hence, $ (r - s)^2 = r^2 - 2rs + s^2 = 6, $ so $ |r - s| = \boxed{\sqrt{6}} $.