$2^{2}+1=5$, $2^{8}+1=257$.  $2^{3}=8$, $3^{3}=27$, $4^{3}=64$, $5^{3}=125$, $6^{3}=216$.  Thus, the answer is $5$. 
How many perfect cubes lie between $2^{2}+1$ and $2^{8}+1$, inclusive? (Adapted from 2018 AMC 8 Problem, Question \#25) 
  - $$2$$ 
  - $$3$$ 
  - $$4$$ 
  - $$5$$ 
  - $$6$$ 