Check if there is only one simple path in graph between nodes x and y`

Let's say we have given simple undirected graph $G$ having $N$ nodes and $M$ bidirectional edges. For given $x$ and $y$ we want to check if in the graph there is only one simple path between them.

What I was thinking for, is that we should find all the cycles in the graph and run bfs from $x$ to $y$, if the path doesn't move through the vertices that are in some cycle, there is only one path, and otherwise there are more.

Is there some other way to check this?

Raphael's suggestion to use 2-shortest path is correct but more complex than necessary.

Your approach is unfeasible, the number of cycles is factorial in the size of input.

Here's a sketch of a solution in time $O(|V|+|E|)$.

Find a path $p_1, \dots p_n$ from $x$ to $y$ in $O(|V|+|E|)$. If no such path exists, answer $\textsf{NO}$. Otherwise, mark all the nodes in the path. That path is indeed unique if and only if no node in that path reaches a node that comes after it. Starting from the first node of the path $p_1$, pick a random edge going out from $p_1$ other than the one that connects it to $p_2$ and start a DFS from there. If you reach a marked node, answer $\textsf{NO}$, if you never reach a marked node, cut that branch from the graph and pick another random outgoing edge. Continue in a similar fashion until all outgoing edges are exhausted, then start again with the following node until you reach $p_n$.
If you finish the procedure without ever answering $\textsf{NO}$, answer $\textsf{YES}$. The entire procedure has the cost of a DFS, $O(|V|+|E|)$, therefore the overall time is $O(|V|+|E|)$.

StackExchange Computer Science Q#85810, answer score: 5