2016 year is still messing with new problems

Clearly given is the expansion of $(1+1)^{2016}$ i.e $\Large \sum_{ i=0 }^{2016} \binom{2016}{i} =(1+1)^{2016}$ and this is $x^y$ when |x-y| is least possible, this is the case when $\\2^{2016}=(2^8)^{252}=256^{252}$ Hence $\huge 256+252=\boxed{ \color{#D61F06}{\boxed{508}}}$

Consider the equation $(1+x)^{2016} = \sum_{n=0}^{2016} \binom{2016}{n} 1^{2016-n}x^n$

Substituting $x=1$ gives $2^{2016} = \sum_{n=0}^{2016} \binom{2016}{n}$

What we require, however, is to make the base and exponents have the least differences possible. Since $2016 = 8\times252$ and $2^8 = 256$ we can put $2^{2016}$ to $(2^8)^{252} = 256^{252}$ . Hence $x+y = 256+252=508$