Simplifying triangulations of 3-manifolds

There are many such examples, depending upon what you mean by "triangulation". If a triangulation is just a glueing of tetrahedra along faces, then the simplest one is probably the following: the 3-sphere has a triangulation with 1 tetrahedron, and a triangulation with 2 tetrahedra (it is a nice excercise to find them). However, you cannot apply any $4 \to 1$ or $3 \to 2$ move to a triangulation having only 2 tetrahedra.

If by "triangulation" you mean a "simplicial triangulation", then it suffices to construct a triangulation of the 3-sphere such that every edge meets at least 4 distinct tetrahedra. In this case, no $4 \to 1$ and no $3 \to 2$ moves can apply.

For instance, you can take a triangulation of the 2-sphere such that every vertex is incident to at least 4 triangles (as an example, the octahedron). By doing a suspension of this triangulation you get a triangulation of the 3-sphere such that every edge is incident to at least 4 tetrahedra.

If I were to bet on the question, I would expect that expanding and contracting moves would be needed. My guess to construct an example would be to take an unknot that required a type II move that increased the number of crossings, and follow that simplification. To get a 3-mfd from such an example, try a Dehn filling of the knot (or better just work with the manifold with boundary). I think that it is not too hard to write down a triangulation from the diagram (several CW complexes and handle decompositions are apparent). Examine what the Reidemeister moves do to the triangulation, and decompose each in terms of Pachner moves.

For example the number of 2-handles differs on the two sides of a type II move.

I certainly haven't worked out the details, but it should now be a good exercise in handle/cell/triangulation moves.