Does the pattern of identities for $(x+y+z)^{2n-1}$ continue past $2 n - 1 = 5$? We have the following identities.
\begin{align}
(x+y+z)^3&=x^3+y^3+z^3+3(x+y)(y+z)(z+x)\\
(x+y+z)^5&=x^5+y^5+z^5+5(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx)
\end{align}
I'm wondering is there any pattern like above in the expansion of $(x+y+z)^{2n-1}$ where $n\in\mathbb{Z}^+$? The pattern I see so far is appearance of the coefficient $2n-1$ and $(x+y)(y+z)(z+x)$.

The above expansions can be found by factoring $(x+y+z)^{t}-x^{t}-y^{t}-z^{t}$ by using Cyclic Polynomials. For example for $t=5$ the factorization is provided on Brilliant.

I tried factoring $(x+y+z)^{7}-x^{7}-y^{7}-z^{7}$ using the method I got:
$$(x+y)(y+z)(z+x) [A(x^4+y^4+z^4)+B(x^3y+y^3z+z^3x)+C(x^2y^2+y^2z^2+z^2x^2)]$$
Here I'm not sure whether I should include $D(xy^3+yz^3+zx^3)$ too or not.

Denote $$\Delta_m(x, y, z) := (x + y + z)^m - x^m - y^m - z^m .$$
For odd $m$, $\Delta_m(x, y, y) = 0$ so $y + z$ divides $\Delta_m(x, y, z)$; by symmetry so do $z + x$ and $x + y$. Hence, $$(x + y + z)^m = x^m + y^m + z^m + (y + z)(z + x)(x + y) q_m(x, y, z)$$ for some symmetric, homogeneous polynomial $q_m$ of (even) degree $m - 3$. One can always write a symmetric polynomial as a polynomial in elementary symmetric polynomials, in this case,
$$1, \quad x + y + z, \quad y z + z x + x y, \quad x y z .$$
In general, the coefficient of $x^i y^j z^k$, $i + j + k = m$, in the expansion of $(x + y + z)^m$ is the multinomial coefficient ${m \choose{i, j, k}} = \frac{m!}{i! j! k!} .$ If $m = p$ is prime, then except when one of $i, j, k$ is equal to $p$, corresponding to the monomials $x^p, y^p, z^p$, respectively, we have $i, j, k < p$ and so $p \mid {p\choose{i, j, k}}$. Thus $p \mid \Delta_p(x, y, z)$ (over $\Bbb Z$), and hence also $p \mid q_p(x, y, z)$.
Example ($m = 7$) By our previous observation, $7 \mid q_7(x, y, z)$ because $7$ is prime, and because $\deg q_7 = 4$ we get one potential summand for each (unordered) partition of $4$ into sums of positive integers no larger than $3$ (namely, $[1,1,1,1]$, $[2,1,1]$, $[2,2]$, $[3,1]$). More precisely,
\begin{multline}
q_7(x, y, z) = 7[A (x + y + z)^4 + B (y z + z x + x y) (x + y + z)^2 \\ + C (y z + z x + x y)^2 + D x y z (x + y + z)]
\end{multline}
for some integer coefficients $A, B, C, D$. Specializing the above formulae to $z = -y$ and comparing like coefficients immediately gives $A = C = 1$ and $B + D = -1$. Then instead specializing to $y = x, z = 0$ gives $B = -2$ and thus $D = 1$.
Alternatively, we can decompose
\begin{multline}
q_7(x, y, z)
    = 7 [
           E (x^2 y z + y^2 z x + z^2 x y)
         + F (y^2 z^2 + z^2 x^2 + x^2 y^2) \\
         + G (y^3 z + y z^3 + z^3 y + z y^3 + x^3 y + x y^3)
         + H (x^4 + y^4 + z^4) ] ,
\end{multline}
and again solving for the coefficients gives $E = 5, F = 3, G = 2, H = 1$.
For general odd $m$, for both decompositions of $q_m$ there are $\left[\frac{m^2}{12}\right]$ terms, where $[\,\cdot\,]$ denotes rounding to the nearest integer; see A001399.
Counterexample ($m = 9$) As alex.jordan pointed out in the comments, if $m$ is composite it need not divide $q_m(x, y, z)$, and Oscar Lanzi observed in a comment replying to the other answer that's already the case for $m = 9$, the smallest odd composite number: Indeed the coefficients of the monomials $x^i y^j z^k$ of $q_9(x, y, z)$ turn out to be $9, 27, 57, 69, 105, 189, 273$, some of which are not divisible by $9$, so $9 \nmid q_9(x, y, z)$
Probably one could use Newton's Identities to derive an explicit formula for $q_m$ in terms of $m$. In the standard notation $e_i(x, y, z)$ for elementary symmetric polynomials and $p_i(x, y, z)$ for power sums, $(y + z)(z + x)(x + y) = e_1 e_2 - e_3 = \frac{1}{3} (p_1^3 - p_3)$, so your observation is that for odd $m$ the difference $e_1^m - p_m$ factors as $$e_1^m - p_m = (e_1 e_2 - e_3) q_m = \frac{1}{3} (p_1^3 - p_3) q_m$$ for some polynomial $q_m$. I don't know formulae offhand, but Bell numbers might play a role.
