Prove that if x,y,z are positive integers such that $x^3+y^3+z^3=3(x+y+z+xyz)$ then they must be consecutive numbers

Use the well-known identity $$x^3+y^3+z^3-3xyz = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right)$$ to have, from your given equation (since the left hand side equals $3(x+y+z)$, $$\begin{aligned} 3(x+y+z) & = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right) \\ \implies 6 &= (x-y)^2+(y-z)^2+(z-x)^2 \ (\because x,y,z>0 \implies x+y+z\ne 0) \end{aligned}$$ also, $x,y,z$ being positive integers, the squares in the last line $(x-y)^2, (y-z)^2, (z-x)^2$ are perfect square integers, and the only way three perfect squares add up to $6$ is $$6=1+1+4$$ so, two of the differences $|x-y|$ and $|y-z|$ (without loss of generality) must be $1$, naturally the $|x-z|$ must be $2$.

First, subtract $3xyz$ from both sides of the equation: $$x^3+y^3+z^3=3(x+y+z+xyz) \\ x^3+y^3+z^3-3xyz = 3(x+y+z)$$ Now, using this factorisation, observe: $$ (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=3(x+y+z)$$ Since $x,y$, and $z$ are positive integers which implies $x+y+z\neq 0$, you can divide the equation by $x+y+z$ to get $$ x^2+y^2+z^2 - xy-yz-zx = 3$$ Now, from the above equation, it's easy to observe that $$(x-y)^2+(y-z)^2+(z-x)^2=6$$

Now, if two of the differences in the brackets are greater than $1$ then the sum will be greater than $2^2 + 2^2 = 8 > 6$ (contradiction!).

Also, if all are equal to $1$, then the sum will be $3 \neq 6$ (contradiction!).

Hence, one of the differences is $2$ and the other two are equal to $1$ (so that $2^2+1^2+1^2=6$ as desired) which implies $x,y$, and $z$ are consecutive integers.

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Why is that? Now goes the explanation:

Since you have already shown that one of the differences is $2$ and the other two are $1$, without loss of generality, you may take $$ \begin{cases} x - y = 1 \\ y - z = 1 \\ x - z = 2 \end{cases} \implies \begin{cases} x = y + 1 \\ y = z + 1 \\ x = z + 2 \end{cases} $$ Hence, $$(x,y,z) = (z+2, ~z+1, z)$$