Let $ y = \frac{x^2 + 4x}{x - 1} $. Then we can write the given equation as \[y + \frac{72}{y} - 18 = 0,\]so $ y^2 - 18y + 72 = 0 $. This factors $ (y - 6)(y - 12) = 0, $ so $ y = 6 $ or $ y = 12 $. If $ \frac{x^2 + 4x}{x - 1} = 6, $ then $ x^2 + 4x = 6x - 6, $ or $ x^2 - 2x + 6 = 0 $. This quadratic has no real solutions. If $ \frac{x^2 + 4x}{x - 1} = 12, $ then $ x^2 + 4x = 12x - 12, $ or $ x^2 - 8x + 12 = 0 $. This factors as $ (x - 2)(x - 6) = 0, $ so the solutions are $ \boxed{2,6} $.