Triples!

$2018^2 + b^2 = c^2 \Rightarrow c^2 - b^2 = 2018^2 \Rightarrow (c-b)(c+b) = 2^2\cdot 1009^2$ .

Since $c-b < c+b$ and $c$ and $b$ must be positive integers, there are only four possibilities:

(1) $(c-b, c+b) = (1, 4\cdot 1009^2)$

(2) $(c-b, c+b) = (2, 2\cdot 1009^2)$

(3) $(c-b, c+b) = (4, 1009^2)$

(4) $(c-b, c+b) = (1009, 4\cdot 1009)$

Since $c-b$ and $c+b$ must have the same parity, only case (2) has an integer solution; therefore, the answer is $\boxed{1}\,$ .