Imagine taking a circle $C$ which is large enough such that (i) all of the intersection points between lines in the arrangement lie inside $C$ and (ii) every line intersects $C$ twice.
It follows from (i) that bounded regions do not intersect $C$ (since all their vertices are inside $C$). Furthermore, any unbounded region must intersect $C$ (The region must have points outside $C$ since it's unbounded. Take a point on the boundary of the region lying outside $C$, follow the boundary line towards $C$ -- it can't intersect another boundary before hitting $C$ by (i))
Now imagine making a single circuit travelling around $C$. There will be $2n$ distinct points at which $C$ intersects one of the lines (since each line intersects twice and (i) implies that no two lines intersect $C$ in the same place). So we only change regions $2n$ times before returning back where we started, implying we've passed through at most $2n$ regions total. So the number of unbounded regions is at most $2n$.
Equality holds so long as we don't encounter the same region twice as we travel around the circle (which I believe only happens if we have $n$ parallel lines). For example, we get equality if we just take $n$ concurrent lines. The number of regions is $2n$ (very far from maximal), but they're all unbounded.