By the properties of logarithms, we can rearrange the equation to read $x^{2017}=2017x$ with $x=\log_b a$. If $x\neq 0$, we may divide by it and get $x^{2016}=2017$, which implies $x=\pm \root{2016}\of{2017}$. Hence, we have $3$ possible values $x$, namely
$$x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad x=-2017^{\frac1{2016}}.$$
Since $\log_b a=x$ is equivalent to $a=b^x$, each possible value $x$ yields exactly $199$ solutions $(b,a)$, as we can assign $a=b^x$ to each $b=2,3,\dots,200$. In total, we have $3\cdot 199=\boxed{\textbf{(E) } 597}$ solutions.