Solve in integers $b^{11}-1=a^{2016}+a^{2015}+\dots+1$ 
Find all integers $(a,b)$ satisfying $$b^{11}-1=a^{2016}+a^{2015}+\dots+1.$$

Obviously, we can get the factorisation $(b-1)(b^{10}+\dots+1)=a^{2016}+a^{2015}+\dots+1$, but I'm not sure how to proceed from there. 

Note that $a^{2016}+a^{2015}+\dots+a+1$ has only prime factors that are $0,1$ modulo $2017$. 
This implies that $b^{11} \equiv 1,2 \pmod {2017}$.
Therefore, $b^{2013} \equiv 1,2^{183} \pmod {2017}$ or $b^3,2^{183}b^3 \equiv 1 \pmod {2017}$.
This implies that $b \equiv 1 \pmod {2017}$ or $b^2+b+1 \equiv 0 \pmod {2017}$ or $2^{61}b \equiv 1 \pmod {2017}$ or $2^{122}b^2+2^{61}b+1 \equiv 0 \pmod {2017}$. 
Since $b^{10}+b^9+\dots+b+1 \equiv 0,1 \pmod {2017}$, it is not difficult to see that $b \not \equiv 1 \pmod {2017}$ and $b^2+b+1 \not \equiv 0 \pmod {2017}$. 
Also, note that if $2^{61}b \equiv 1 \pmod {2017}$ or $2^{122}b^2+2^{61}b+1 \equiv 0 \pmod {2017}$, $b-1 \not \equiv 1,0 \pmod {2017}$. Therefore, no such solutions exist.    
