Let $a=x+y+z$, $b=xy+yz+xz$, and $c=xyz$. We have $b=-4$ and $a^2-2b=x^2+y^2+z^2=24$, so $a=\pm4$. The last equation is $a(a^2-3b)+6c=16$, so $c=\frac{8\mp56}3$. Case 1: $(a,b,c)=(4,-4,-16)$ We have that $x,y,z$ are the roots of $x^3-4x^2-4x+16=0$, so $(x,y,z)=(-2,2,4)$ and permutations. Case 2: $(a,b,c)=\left(-4,-4,\frac{64}3\right)$ Since $a,b,c\in\mathbb Z$, we have that $abc\in\mathbb Z$, contradiction.

jasperE3 wrote: Let $a=x+y+z$, $b=xy+yz+xz$, and $c=xyz$. We have $b=-4$ and $a^2-2b=x^2+y^2+z^2=24$, so $a=4\sqrt2$. It's impossible for the sum of three integers to not be an integer, hence no integral solutions. Not so fast. $a^2-2b=24 \Leftrightarrow a^2=16$, not $32$. Solution: As above said, $b=4$. Also $a=\pm 4$. Now because of the 3rd equality $a(a^2-3b)+6c=16$ if $a=-4$, then we get that $c=\frac{32}{3}$, but this is impossible, since $a,b$ and $c$ are integers. Thus $a=4$ and $c=-16$, so we have $x+y+z=4$; $xy+xz+yz=-4$; $xyz=-16$ . Now we construct a polynomial with roots $x$, $y$ and $z$, i.e $(t-x)(t-y)(t-z)=0\Leftrightarrow t^3-(x+y+z)t^2+(xy+xz+yz)t-xyz=0\Leftrightarrow$ $t^3-4t^2-4t+16=0\Leftrightarrow (t-2)(t+2)(t-4)=0\Rightarrow$ $(x,y,z)=(2,-2,4)$ and permutations.

@2above I'm not sure that $a^2-2b=24$ with $b=-4$ gives that $a=4\sqrt{2}$. In fact, it follows that $a=\pm 4$. If $a=-4$, substituting in the third equation we don't get an integral $c$ which is a contradiction. If $a=4$, substituting in the third equation we can easily see that $c=-16$. By Vieta, $x, y, z$ must be roots of the polynomial $t^3-4t^2-4t+16=(t-4)(t+2)(t-2)$ and we conclude that the solutions stated by @below are indeed the only (integral) ones. Sorry for posting what is essentially the same sol @above and @2above, I didn't notice you had posted yours before me.

Through guessing, $(4,2, -2)$ and permutations work. We also know that there can only be one set of three integers $\{x,y,z\}$ that works (this follows from say the fact that we can construct a cubic polynomial with $x, y, z$ as roots)

first of all we have $(x+y+z)=\pm 4$ also we have $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(\pm 4)(24+4)$ and also we have $x^3+y^3+z^3+3xyz=16$ so we have $6xyz=4(4-(\pm 28))$ , since $xyz \in \mathbb{Z}$ we must have $xyz=-16$ also since we have $x^2+y^2+z^2=24$ this forces $(x^2,y^2,z^2)=(4,4,16)$ and their permutations. W.L.O.G $x= \pm 2 , y= \pm 2 , z=\pm 4$. If all are negative we can't have $\sum xy =-4$ , so in order to have $\sum xy =-4$ we must have two of them to be positive and one negative , which gives only possible pair is $(2,-2,4)$ so in conclusion we have all solution that works is $\boxed{(2,-2,4)}$ and their permutations. $\blacksquare$