Ordered pairs of factors

$ab = 4^{2017} = (2^2)^{2017} = 2^{4034}$. The number of positive factors of $2^{4034}$ is $4034 + 1 = 4035$.

Since $4035$ is odd, $2^{4034}$ is a perfect square, so $a = b = 2^{2017}$ is one solution. For pairs with $a \neq b$, half of the remaining $4034$ factors satisfy $a < b$.

Total pairs $(a,b)$ with $a \le b$ = $\frac{4035 - 1}{2} + 1 = 2017 + 1 = 2018$.