2018 in all bases

We seek positive integers $a,b$ , with $b>2$ , such that $ab^3+2b^2+2=2018$ or $ab^3+2b^2=2016=2^5 3^2 7$ . Thus $b^2$ must divide $2016$ ; the only candidates are $b=3,4,6,12$ . We check that all of those "work"; thus the answer is $\boxed{4}$ .

We can check in small bases and find that

$2018_{10} = 2202202_3 = 133202_4 = 13202_6 = 1202_{12}$

Since the numbers keep getting shorter with every base, after $1202$ there is only the possibility $202$ left. This means that

$2 \cdot b^2 + 0 \cdot b^1 + 2 \cdot b^0 = 2018 \Leftrightarrow b = \pm 7 \sqrt{12} \notin \mathbb{N}$

So, $b = 3,4,6,12$ are the 4 only possibilities.