Discard the existence of a 720-cell as a regular 4-polytope It is possible to quickly dismiss the existence of a 720-cell as a  regular 4-polytope ?
Only with the following structural elements :
720 cell - 120 faces - 120 edges - 720 vertices
Similar to the 24 cell but, with the Icosidodecahedron and Rhombic triacontahedron as 3d "analogues"
The answer not necessarily has to be with purely geometrical methods.
I had this question time ago, but I forgot to ask it. 
 A: A regular $4$-polytope must consist of identical cells, each of which is a regular polyhedron, with an equal number of cells (at least three) meeting at each edge. This gives only six possible regular $4$-polytopes, all have been thoroughly investigated, none has $720$ cells. 
You appear to be thinking about something less regular, however; neither the icosidodecahedron nor the rhombic triacontrahedron is a regular polyhedron, though they each have many symmetries.
One possibility is as follows. First, note that a bipyramid is formed by a pyramid and its mirror image sharing a base. In a pentagonal bipyramid, each pyramid has a pentagonal base, hence the resulting polyhedron has ten triangular faces and seven vertices.
For each of the $720$ pentagonal faces of a regular $120$-cell $4$-polytope, construct a pentagonal bipyramid crossing the face and using the five vertices of the face as five of the vertices of the bipyramid.
Set the proportions of the triangular faces such that the $12$ bipyramids on the faces of a dodecahedral cell meet at a common vertex.
This is an analogy to the construction of a rhombic triacontrahedron by constructing a rhombus across each edge of a dodecahedron, proportioning the shape of the rhombus so that the five rhombi that cross the edges of one face of the dodecahedron all meet at a common vertex.
The result is not a regular polytope, because the pentagonal bipyramid is not a regular polyhedron.
I have been unable to find an accepted name for this polytope,
but it has been described by multiple authors as the dual of a rectified $600$-cell $4$-polytope (which is another way of constructing the same figure I described above).
It is mentioned on the Wikipedia page "bipyramid".
These numbers are not all achievable, however:

720 cell - 120 faces - 120 edges - 720 vertices

Note that a cell must have at least four faces, but a face is never the face of more than two cells. Hence the number of faces must always be at least twice the number of cells.
It is not possible to have a $720$-cell convex $4$-polytope with fewer than $1440$ faces.
If there are $C$ cells and every cell has $N$ faces then there must be $CN/2$ faces.
Similarly, at least four edges must meet at every vertex, but each edge can meet only two vertices, so there must be at least twice as many edges as vertices. A convex $4$-polytope with $720$ vertices must have at least $1440$ edges.
So we can reject the hypothesis of $120$ faces and $120$ edges immediately.
With ten faces per cell, this $720$-cell polytope must have $720\times 10/2 = 3600$ faces.
All the edges of the $120$-cell are edges of the $720$-cell, which is $1200$ edges, and in addition there is an edge from the center of every cell of the $120$-cell to each of the $20$ vertices of that cell, which is an additional $120\times20 = 2400$ edges, for a total of $3600$ edges.
The dual of this polytope also has $720$ cells, $3600$ faces, $3600$ edges, and $720$ vertices, but the dual is not the same polytope (unlike the $24$-cell, which is its own dual).
