We start with a claim. Claim. For any $t\in\mathbb{N}$, there exists primes $p_1<p_2<\cdots<p_t$ such that $\textstyle p_i\nmid \prod_{1\le j\le t}(p_j^3+1)$. Proof of Claim. We prescribe primes inductively, where $p_i\equiv 5\pmod{6}$ at each step. Note that $p_i\nmid p_j^2-p_j+1$: otherwise $p_i\mid (2p_j-1)^2+3$, and $(-3/p_i)=1$, which is impossible as $p_i\equiv 5\pmod{6}$. Now, having chosen $p_1>3$, choose $p_2$ such that $p_2\equiv 5\pmod{6}$ and $p_2\not\equiv -1\pmod{p_1}$. Continuing, choose $p_j$ such that $p_j\equiv 5\pmod{6}$ and $p_j\not\equiv -1\pmod{p_i}$ for $1\le i\le j-1$. This completes the claim. Having shown the claim, consider $\textstyle n=2\prod_{1\le j\le t}(p_j^3+1)$, and let $A_j = n/(2(p_j^3+1))$. Consider $(x_j,y_j,z_j,t_j) = (2A_j,A_jp_j^3,A_jp_j^3,A_jp_j^2)$. Clearly, $x+y+z = 2A_j(1+p_j^3)=n$, as desired. Since $p_j\nmid A_j$ for any $j$ (from claim above), distinctness is clear.