Is the longest trail problem easier than the longest path problem?
From the comments above: the Hamiltonian cycle problem remains NP-complete even in grid graphs with max degree 3 [1], but in these graphs every traversal of a node requires two edges and at most one edge remains unused, so a node cannot be traversed twice by an Eulerian path.

So apparently there is an immediate reduction from the Hamiltonian cycle problem to your problem: 
given a grid graph with max degree 3 $G = (V,E)$, just ask for a trail of length $|V|$.

But all three edges of the node at the end of the trail can be used; to avoid this situation you can pick the top-left node $u$ of the grid graph (which has degree two) and add two nodes: $V' = V \cup \{u',u''\}$ and a new edges $E = E \cup  \{(u,u'), (u,u'')\}$ and ask for a trail of length $|V'| = |V|+2$: informally the added edge forces $u',u''$ to be the endpoints of the trail.

[1] Christos H Papadimitriou, Umesh V Vazirani, On two geometric problems related to the travelling salesman problem, Journal of Algorithms, Volume 5, Issue 2, June 1984, Pages 231-246, ISSN 0196-6774