Let $ p, $ $ q, $ $ r $ be positive constants. Then by AM-GM, \begin{align*} a^2 + p^2 &\ge 2pa, \\ b^3 + b^3 + q^3 &\ge 3qb^2, \\ c^4 + c^4 + c^4 + r^4 &\ge 4rc^3.\end{align*}Hence, \begin{align*} a^2 + p^2 &\ge 2pa, \\ 2b^3 + q^3 &\ge 3qb^2, \\ 3c^4 + r^4 &\ge 4rc^3.\end{align*}Multiplying these inequalities by 6, 3, 2, respectively, we get \begin{align*} 6a^2 + 6p^2 &\ge 12pa, \\ 6b^3 + 3q^3 &\ge 9qb^2, \\ 6c^4 + 2r^4 &\ge 8rc^3.\end{align*}Hence, \[6(a^2 + b^3 + c^4) + 6p^2 + 3q^3 + 2r^4 \ge 12pa + 9qb^2 + 8rc^3.\quad (*)\]We want to choose constants $ p, $ $ q, $ and $ r $ so that $ 12pa + 9qb^2 + 8rc^3 $ is a multiple of $ a + b^2 + c^3 $. In other words, we want \[12p = 9q = 8r.\]Solving in terms of $ p, $ we get $ q = \frac{4}{3} p $ and $ r = \frac{3}{2} p $. Also, equality holds in the inequalities above only for $ a = p, $ $ b = q, $ and $ c = r, $ so we want \[p + q^2 + r^3 = \frac{325}{9}.\]Hence, \[p + \frac{16}{9} p^2 + \frac{27}{8} p^3 = \frac{325}{9}.\]This simplifies to $ 243p^3 + 128p^2 + 72p - 2600 = 0, $ which factors as $ (p - 2)(243p^2 + 614p + 1300) = 0 $. The quadratic factor has no positive roots, so $ p = 2 $. Then $ q = \frac{8}{3} $ and $ r = 3, $ so $ (*) $ becomes \[6(a^2 + b^3 + c^4) + \frac{2186}{9} \ge 24(a + b^2 + c^3).\]which leads to \[a^2 + b^3 + c^4 \ge \frac{2807}{27}.\]Equality occurs when $ a = 2, $ $ b = \frac{8}{3}, $ and $ c = 3, $ so the minimum value of $ a^2 + b^3 + c^4 $ is $ \boxed{\frac{2807}{27}} $.