$PSL(3,4)$ has no element of order $15$

The splitting field for the nontrivial irreducible representations of the cyclic group of order 5 over the field $F_4$ of order 4 is $F_{16}$, so all such representations have degree 2 over $F_4$.

So (using Maschke's Theorem), a nontrivial representation of a group of order 5 of degree 3 over $F_4$ must be the direct sum of a nontrivial irreducible of dimension 2 and the trivial module.

The centralizer of the image of this representation in ${\rm GL}_3(4)$ must fix both of these irreducible constituents. The centralizer of the constituent in ${\rm GL}_2(4)$ is cyclic of order 15, and the centralizer of the trivial constituent is just the scalar matrices - i.e. cyclic of order 3. So the full centralizer in ${\rm GL}_3(4)$ has order 45.

Since not all elements in this centralizer have determinant 1, its intersection with ${\rm SL}_3(4)$ has order 15. Since it contains the scalar matrices in ${\rm SL}_3(4)$, it follows that all elements of order 15 in ${\rm SL}_3(4)$ have fifth power equal to a scalar matrix, and so ${\rm PSL}_3(4)$ has no element of order 15.

I have just noticed that I have assumed that the inverse image in ${\rm SL}_3(4)$ of an element of order 15 in ${\rm PSL}_3(4)$ centralizes an element of order 5, but I will leave you to show that.