The other examples are infinitely presented (indeed, generated). That is, they are of the form $G=\langle X; R\rangle$ where $X$ or $R$ is infinite. In group theory, the challenge is often to find finitely presented groups with extreme properties because it is "easy" to find infinitely presented groups (indeed, finitely-generated infinitely-presented groups!) with extreme properties$^{\ast}$. For example, there exist infinite finitely generated groups where every element has finite order, but all known examples are infinitely presented...or hopfian groups - it is easy to reel of a non-Hopfian group, but finding one which is finitely presented is harder (indeed, the other two answers both give you non-hopfian groups). See this question for the definition and examples of non-Hopfian groups.
So, I now need to give you a finitely presented example. A few things we need to know first, and they are all well-known and proofs should all be easily obtainable (they should all be in Magnus, Karrass and Solitar's book "Combinatorial Group Theory", or can be found on this here website, or you can try and prove them yourself - one of them is pretty easy...).
- If a group is finitely presented then every subgroup of finite index is finitely presented,
- Subgroups of free groups are free,
- If $G$ is a free group of rank at least two then $G^{\prime}$ is the free group of countable rank, and is of infinite index.
Okay, let $G=F_n$ be the free group on $n$ generators, $n\geq 2$. Then $G$ has subgroups of finite index (why?) and these are all finitely presented and free. So, take one of them and call it $F_m$ (as it is free, on $m$-generators). On the other hand, take $m$-elements of the basis of $G^{\prime}$. These are both $m$-generated, and as subgroups of free groups are free and free groups are isomorphic if and only if they are free on the same cardinality of generators we are done!
$\ast$: I say "easy", but I really mean "easier". For example, Ol'shanskii proved that there exist infinite groups where every subgroup has order $p$ for some fixed, large enough prime $p$. His construction gives a two-generated, infinitely presented group. He has a book, which is pretty much 500 pages long, and is basically just the proof. It isn't light reading, either (it is called "Geometry of Defining Relations in Groups").
The index of this subgroup is $n=|ad-bc|$. The quickest argument I can think of right now is that $n$ is the total area of the torus $T=\mathbf R^2/\Gamma$ (the parallelogram spanned by $(a,b)$ and $(c,d)$ is a fundamental domain for $\Gamma$), and that the projection $T\to\mathbf R^2/\mathbf Z^2$ is a locally area-preserving covering of a torus of area $1$, so that it must be an $n$-fold covering; the size of the preimage of $(0,0)$ is the index of $\Gamma$ in $\mathbf Z^2$.
As suggested in the comment by Steve D, there is also a way to see this that avoids using real numbers and area considerations. One can start with a matrix
$$
\begin{pmatrix}a&c\\b&d\end{pmatrix}
$$
whose columns express a pair of generators of $\Gamma$ in the standard basis of $\mathbf Z^2$, and repeatedly modify it by either right-multiplying by a matrix of $\mathbf{GL}(2,\mathbf Z)$ (integer entries, determinant $\pm1$) which corresponds to choosing a different pair of generators of $\Gamma$, or left-multiplying by a matrix of $\mathbf{GL}(2,\mathbf Z)$, which corresponds to changing the basis of $\mathbf Z^2$ that the generators are expressed in. These operations do not change the absolute value of the determinant, and one can arrange for the resulting matrix to become diagonal (the Smith normal form is an even more special form which one does not need here, but the theory of Smith normal forms at least shows this is possible). Now each generator is just a multiple of a basis vector of $\mathbf Z^2$, and it is now clear that the index of $\Gamma$ is the absolute value of the product of those multiples (diagonal entries), which is the absolte value of the determinant, which hasn't changed. so it is $|ad-bc|$.
Best Answer
Did you try GAP? It should give you all index 6 subgroups. The group has presentation $\langle a,b \mid a^6=b^4=1, a^3=b^2\rangle.$ In general classification of all finite index subgroups of that group is a hopeless task. There are too many of them. For example every finite group generated by an element of order two and an element of order 3 is a quotient of that group. The kernel is then of finite index. And there are lots of such finite groups.
But one can try to apply something similar to Stallings core graphs that work for the free groups.