For $ (ax + b)(x^5 + 1) - (5x + 1) $ to be divisible by $ x^2 + 1, $ it must be equal to 0 at the roots of $ x^2 + 1 = 0, $ which are $ \pm i $. For $ x = i, $ \begin{align*} (ax + b)(x^5 + 1) - (5x + 1) &= (ai + b)(i + 1) - (5i + 1) \\ &= -a + ai + bi + b - 5i - 1 \\ &= (-a + b - 1) + (a + b - 5)i.\end{align*}Then we must have $ -a + b - 1 = a + b - 5 = 0 $. Solving, we find $ (a,b) = \boxed{(2,3)} $. For these values, \[(ax + b)(x^5 + 1) - (5x + 1) = 2x^6 + 3x^5 - 3x + 2 = (x^2 + 1)(2x^4 + 3x^3 - 2x^2 - 3x + 2).\]