So i have come up with a foldr solution. It should be similar to what @Will Ness has demonstrated but not quite i suppose as we don't need a separate empty list check in this one.
The thing is, while folding we need to encapsulate the previous element and also the state (the result) in a function type. So in the go helper function f is the state (the result) c is the current element of interest and p is the previous one (next since we are folding right). While folding from right to left we are nesting up these functions only to run it by applyying the head of the input list to it.
go :: Ord a => a -> (a -> [a]) -> (a -> [a])
go c f = \p -> let r = f c
in if c > p then c:r else r
greaters :: Ord a => [a] -> [a]
greaters = foldr go (const []) <*> head
*Main> greaters [1,3,2,4,3,4,5]
[3,4,4,5]
*Main> greaters [5,10,6,11,7,12]
[10,11,12]
*Main> greaters [651,151,1651,21,651,1231,4,1,16,135,87]
[1651,651,1231,16,135]
*Main> greaters [1]
[]
*Main> greaters []
[]
As per rightful comments of @Will Ness here is a modified slightly more general code which hopefully doesn't break suddenly when the comparison changes. Note that const [] :: b -> [a] is the initial function and [] is the terminator applied to the result of foldr. We don't need Maybe since [] can easily do the job of Nothing here.
gs :: Ord a => [a] -> [a]
gs xs = foldr go (const []) xs $ []
where
go :: Ord a => a -> ([a] -> [a]) -> ([a] -> [a])
go c f = \ps -> let r = f [c]
in case ps of
[] -> r
[p] -> if c > p then c:r else r