Let us suppose first that all the coefficients in your polynomial $p$ are non-zero.
Consider the set of vectors of $\mathbb R^3$ given by the rows of the following matrix:
7 & 0 & 0 \\
5 & 1 & 0 \\
3 & 2 & 0 \\
1 & 3 & 0 \\
4 & 0 & 1 \\
2 & 1 & 1 \\
0 & 2 & 1 \\
1 & 0 & 2
$$ which comes from the exponents which appear in the monomials in you polynomial $p$.
All of them satisfy the equation $$(\Pi)\qquad\qquad x+2y+3z=7,$$ and this tells us that the convex hull of those eight points is actually a polygon contained in a plane. To be able to make pictures more easily, I will project down to one of the coordinate planes: the projection $\phi:\mathbb R^3\to \mathbb R^2$ on the last two coordinates is injective and affine on that plane (and «it corresponds to looking at $\mathbb R^3$ from above»). The image of our points under this projections are, of course, the rows of
0 & 0 \\
1 & 0 \\
2 & 0 \\
3 & 0 \\
0 & 1 \\
1 & 1 \\
2 & 1 \\
0 & 2
and the convex hull $P_p$ of these in the plane is the following polygon:
We can clearly do this construction from any polynomial $f$ whose exponent vectors lie on a plane parallel to $\Pi$ to obtain a polygon $P_f$.
Suppose now that $p=gh$ is a factorization of your polynomial. An easy consequence of the way polynomials are multiplied implies
Lemma. The exponent vectors of $g$ lie on a plane parallel to $\Pi$ and the same, of course, applies to $h$.
We therefore have two polygons $P_g$ and $P_h$. Now the key observation:
Lemma. We have $P_p=P_g+P_h$.
Here the sum is the so-called Minkowski sum of subsets of the plane.
This last lemma imposes very severe limitations on the shape of the polynomials $g$ and $h$ which can appear in a factorization of $p$.
Indeed, a little thinking will give you the list of all possible pairs of polygons $(A,B)$ whose vertices are points of the plane with non-negative integer coordinates and such that $P_p=A+B$. For each such pair, you get the form of the factos of a possible factorization of $p$.
Now, the factorization you proposed corresponds to the pair of polygons
and in fact this is the only decomposition of $P_p$ as a Minkowski sum (to see this, consider for example the possible heights and widths of the two summands, &c). It follows that assuming the coefficients of $x^7$, $xy^3$, $y^2z$ and $xz^2$ are non-zero, any possible factorization has the shape you found. If one of those coefficients vanishes, then the shape of $P_p$ would be different, and one has to consider more cases—notice that the vanishing of the coefficient of $x^3yz$ does not matter, if the other coefficients are non-zero, because it is in the interior of our polygon.
N.B.: It is easy to see that the polygons corresponding to the shapes of the two factors you found cannot be written as a Minkowski sum. From this it follows that (assuming their coefficients corresponding to the vertices of those polygons do not vanish—and in fact they can't if the same holds for $p$) the two factors you have found are irreducible.