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If $n = 2^{2016}+73$, what is the remainder when $(2^n+1)$ is divided by $(16^{2015}+1)$?

We have $16^{2015}+1 = 2^{8060}+1$. Thus, we need to find $m,r$ where $8060m+r = 2^{2016}+73$ and $0 \leq r < 8060$. This doesn't seem easy, so I was wondering if there was an easier way of solving this question?

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    Let $m=16^{2015}+1 = 2^{8060}+1$. Working $\!\!\pmod{m}$, we have $2^{8060}\equiv -1$. Additionally, $8060=2^2\cdot 5\cdot 13\cdot 31$. It follows that we may compute $2^{2016}+73\pmod{8060}$ by carefully exploiting the CRT. If I am not wrong, we have $2^{2016}+73\equiv 5469\pmod{8060}$, hence: $$ 2^n = 2^{8060k+5469}\equiv (-1)^k 2^{5469}\pmod{m} $$ and since $k$ is odd, $2^n\equiv \color{red}{-2^{5469}}\pmod{m}.$ – 2017-01-03
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    (moved to comments due to a (?) revenge downvote) – 2017-01-03
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    @JackD'Aurizio Is it possible to find the positive remainder here? – 2017-01-03
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    I guess it is $m-2^{5469}$. – 2017-01-03

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