The problem of finding a maximum set with no induced directed triangles is NP-complete via a reduction from maximum independent set.
Let G=(V,E) be an undirected graph and k be an integer, for which we wish to know whether there is an independent set of at least k vertices. Let |V|=n. From G construct a directed graph G'=(V',A'), where V' consists of the disjoint union of V and n additional vertices. For each undirected edge in G create two directed edges between the same vertices in G'. In addition, connect each of the n additional vertices in G' by edges in both directions to the vertices in V, but do not add edges between any two of the n additional vertices.
Then, a triangle-free set of vertices in G' consists either of a subset of V only (with at most n vertices) or it contains some vertices in the set of additional vertices together with an independent set of vertices in G. Therefore, there exists a triangle-free set of size n+k in G' iff there exists an independent set of size k in G.
However, this reduction produces many 2-cycles, so it leaves open the possibility that the problem might be easier in the case that the input graph has no 2-cycles, which you say is the case for your inputs. Unfortunately this special case is still hard: a more complicated reduction based on the same idea shows that the problem remains NP-complete even for 2-cycle-free graphs. In the more complicated reduction, add mn additional vertices rather than simply n. Replace each undirected edge in G by a directed edge, oriented arbitrarily. For each directed edge uv connecting two vertices of V, make n additional vertices in V', each of them connected to u and v to form a directed triangle. Then the resulting graph has mn+k vertices in a triangle-free set iff the original graph has k vertices in an independent set.