Let $ z = x + yi, $ where $ x $ and $ y $ are real numbers. Since $ |z| = 1, $ $ x^2 + y^2 = 1 $. Then \begin{align*} |1 + z| + |1 - z + z^2| &= |1 + x + yi| + |1 - x - yi + x^2 + 2xyi - y^2| \\ &= |(1 + x) + yi| + |(1 - x + x^2 - 1 + x^2) + (-y + 2xy)i| \\ &= |(1 + x) + yi| + |(-x + 2x^2) + (-y + 2xy)i| \\ &= \sqrt{(1 + x)^2 + y^2} + \sqrt{(-x + 2x^2)^2 + (-y + 2xy)^2} \\ &= \sqrt{(1 + x)^2 + y^2} + \sqrt{(-x + 2x^2)^2 + y^2 (1 - 2x)^2} \\ &= \sqrt{(1 + x)^2 + 1 - x^2} + \sqrt{(-x + 2x^2)^2 + (1 - x^2) (1 - 2x)^2} \\ &= \sqrt{2 + 2x} + \sqrt{1 - 4x + 4x^2} \\ &= \sqrt{2 + 2x} + |1 - 2x|.\end{align*}Let $ u = \sqrt{2 + 2x} $. Then $ u^2 = 2 + 2x, $ so \[\sqrt{2 + 2x} + |1 - 2x| = u + |3 - u^2|.\]Since $ -1 \le x \le 1, $ $ 0 \le u \le 2 $. If $ 0 \le u \le \sqrt{3}, $ then \[u + |3 - u^2| = u + 3 - u^2 = \frac{13}{4} - \left( u - \frac{1}{2} \right)^2 \le \frac{13}{4}.\]Equality occurs when $ u = \frac{1}{2}, $ or $ x = -\frac{7}{8} $. If $ \sqrt{3} \le u \le 2, $ then \[u + u^2 - 3 = \left( u + \frac{1}{2} \right)^2 - \frac{13}{4} \le \left( 2 + \frac{1}{2} \right)^2 - \frac{13}{4} = 3 < \frac{13}{4}.\]Therefore, the maximum value is $ \boxed{\frac{13}{4}} $.