If you're using bcrypt, then in addition to the plaintext/password you're hashing, bcrypt requires a salt and a work factor. I'm assuming you're keeping the work factor constant, but you didn't say about the salt.
Generally, the collision probability of a good hash function depends on the size of the hash's output. The birthday paradox makes collisions much more frequent than your intuition allows for; a (very rough) approximation is to take the square root of the number of possible outputs (call the result n) and assume that you'll see a collision after hashing n inputs.
So, if you kept the salt constant, you'd get 184 bits of output space, translating to about 1 collision in 292 hashed inputs. If you used a random salt on every input, you'd have an additional 128 bits of output space, further improving collision resistance.
Note that these are probabilities, not guarantees. So you might produce your first collision much earlier than expected, even though it's unlikely. Still, you should probably plan for collisions. If you actually produced a collision, you could change the salt and try again until you went collision-free; that's simple enough to not complicate your code much.
This article gives you the math behind calculating hash collision probabilities.
