Prove that $\vert PSL(4,2) \vert = \vert PSL(3,4) \vert $ but $PSL(4,2) \ncong PSL(3,4)$.
Attempt: I have shown that the orders of the two groups are equal. Then, consider the set of unitriangular matrices in each. They each form a subgroup, furthermore, they are actually the Sylow 2-subgroups in each. This too, is easy to show. The finisher would be to show that the centers of these two Sylow 2-subgroups are not the same, hence they cannot be isomorphic.
How do I show this?
Your intuition is the right one.
In general, one can prove that the center of the unitriangular matrix group $UT(n,F)$ over a field $F$ is the group of unitriangular matrices $(a_{i \ j})$ such that $a_{i \ j}=0$ for $j>i$ except for $a_{1 \ n}$ which can be any elements of the field. Hence the cardinal of the center is equal to the one of the field.
The proof can be made using elementary matrices. Suppose that $A=(a_{i \ j})$ is a matrix of the center. Then considering the products $L_{i,j}A$ and $A L_{i,j}$ that have to be equal, you can prove that $a_{k \ l}=0$ for $l > k$ and $(k,l) \neq (1,n)$. Conversely, you can prove that a matrix satisfying those conditions commutes with all the unitriangular matrices.
Now the conclusion comes easily as one center has order $2$ while the other one has order $4$.