If we are looking for subrings of $\mathbb Q$ which are not fields, clearly the integers $\mathbb Z$ comes to mind as a first example. The integers are a subring, but not a field since not every non-zero element is invertible.
I'm looking for another example of a subring of $\mathbb Q$ which is not isomorphic to $\mathbb Z$.
I believe that any finite subring of a field is in fact a subfield, so, if I'm correct, we need a non-finite subring that is non-isomorphic to $\mathbb Z$.
First, a couple quick comments.
You are right that any finite subring of a field is also a field. Luckily, this isn't an issue we need to worry about here - $\mathbb{Q}$ has no finite subrings.
(Actually, this depends on the definition of "subring" here - sometimes the trivial subring is allowed, in which case $\mathbb{Q}$ has exactly one finite subring. But meh.)
Moreover, note that $\mathbb{Q}$ is a "minimal" field in that it has no proper subfields (it also embeds uniquely into every field of the same characteristic - this isn't relevant here, but it's neat). So really we can rephrase your question as:
OK, so how do we find proper subrings of $\mathbb{Q}$? Well, they certainly have to contain $\mathbb{Z}$: they have to contain $1$, which gives all of $\mathbb{Z}$. So we're looking at "small" extensions of $\mathbb{Z}$ which get us partway, but not all the way, to $\mathbb{Q}$.
The way we get $\mathbb{Q}$ from $\mathbb{Z}$ is intuitively by adding "$1\over n$" for every $n\in\mathbb{Z}$ (other than zero of course). This suggests a natural "partway" version:
Formalizing the intuition above is a good exercise - and in particular, you'll see that primes play a very convenient role. And a more intricate exercise is to turn the above into an exact characterization of the subrings of $\mathbb{Q}$.