If it's just find, x=21,y=15,z=18 should work.

Note that you can add $1$ or subtract $1$ from $x,y,z$ to change the RHS by a multiple of $3$ while keeping the LHS constant. Infinite solutions, too lazy to find general form.

Let $a=y-z$, $b=x-y$. Then the equation can be written as $-ab(a+b)=3z+2a+b$ or, equivalently $z=-\frac{ab(a+b)+2a+b}{3}$. Easy casework $\mod 3$ shows that $ab(a+b)+2a+b$ is divisible by 3 iff $a \equiv b \equiv 0 \mod 3$. Thus, we can write $a=3m, b=3n$ and get $z=-9mn(m+n)-2m-n$ and hence $y=-9mn(m+n)+m-n$ and $x=-9mn(m+n)+m+2n$. These are all the solutions to the equation. Now, we are left to derive conditions on $m,n$ for $x,y,z$ to be positive. To do this, we introduce $k=-m-n$. Now $k+m+n=0$, $z=9kmn+k-m, y=9kmn+m-n$ and $x=9kmn+n-k$. Since $x+y+z=27kmn$ must be positive, we need $k,m,n \ne 0$ and an either 0 or 2 of the 3 variables to be negative. Since $k+m+n=0$, two of them must be negative and one positive, w.l.o.g. $k>0, m,n<0$. Now, $z=9kmn+k-m>9kmn>0$. We also have $y=9kmn+m-n>9kmn+m=m(9kn+1)$. But $9kn$ is negative and divisible by 9 and hence at most $-9$ thus $9kn+1<0, m<0$ and hence $y>m(9kn+1)>0$. Assume $-n \ge k$. Then $x=9kmn+n-k \ge 9kmn+2n=n(9km+2)$ and as above this implies $x$ must be positive. Assume $-n \le k$. Then $x=9kmn+n-k \ge 9kmn-2k=k(9mn-2)$ and since $9mn \ge 9 >2$ this also forces $x$ to be positive. We finally conclude that for all $k,m,n \in \mathbb{Z}$ with $k+m+n=0$ and $kmn>0$ the numbers $x=9kmn+n-k, y=9kmn+m-n$ and $z=9kmn+k-m$ are solutions to the equation. As explained above, also every solution can be written and this form and hence we have found a complete characterization of the solution set.