An edge $e=uv$ is a bridge in $G$ if there is no path between $u$ and $v$ in $G-e$.
In general, $u$ and $v$ are cut vertices, but there are some special cases you must treat with care.
(1) If $G \simeq K_2$, then depending on the definition of connectivity, the $G-u \simeq K_1$ might and might not be considered connected*.
(2) There is another case in which you would not consider $u$ to be a cut vertex; An edge $uv$ is called a pendant edge if $\deg(u) = 1$. In this case, if $uv$ is a pendant edge and $\deg(v)>1$, you would not typically call $u$ a cut vertex.
All that being said, definitions in graphs are hard (case in point: Diestel). When we try to make definitions very general, special cases sometimes become absurd.
* To address David Richerby's objection, $K_1$ is considered disconnected when deriving connectivity from k-connectivity, in which you require the graph to have more than $k$ vertices. By that definition ("a 1-connected graph is called connected"), $K_1$ is disconnected by having too few vertices.
To answer Thinker's follow-up question; since you wanted to know what to answer to a hypothetical exam question, I would do it this way:
Let $e = uv$ be a bridge. We have three cases:
- $\deg(u), \deg(v) > 1$. The statement is true.
- $\deg(u) = 1, \deg(v) > 1$. The statement is true, $e$ is a pendant.
- $\deg(u) = \deg(v) = 1$. The statement is true if and only if $K_1$ is considered connected.