We have given three distinct nonzero integers $x,y,z$ satisfying $(1) \;\; [x,y] - [x,z] = y - z$. Using the formula $[a,b] \cdot (a,b) = ab$ we can express (1) as $(2) \;\; \frac{xy}{(x,y)} - \frac{xz}{(x,z)} = y - z$. Let $d_1 = (x,y)$ and $d_2 = (x,z)$. Then there exists four nonzero integers $s,t,u,v$ with $(s,t) = (u,v) = 1$ s.t. $(x,y) = (sd_1,td_1)$ and $(x,z) = (ud_2,vd_2)$, which inserted in (2) result in $std_1 - uvd_2 = td_1 - vd_2$, i.e. $(3) \;\; (s - 1)td_1 = (u - 1)vd_2$. The fact that $x = sd_1 = ud_2$ (implying ${\textstyle \frac{s}{u} > 0}$ since $d_1,d_2 > 0$) combined with (3) give us $(4) \;\; (s - 1)tu= (u - 1)sv$. Assume $s \neq 1$. Then $u \neq 1$ by (4). Futhermore $u | (u - 1)sv$ and $s | (s - 1)tu$ by (4), yielding $u|s$ and $s|u$ since $(u,u-1) = (s,s-1) = (s,t) = (u,v) = 1$. Hence $u = s$ (since ${\textstyle \frac{s}{u} > 0}$), yielding $t=v$ by (4), which again implies $d_1=d_2$ by (3). Consequently $y=td_1 = vd_2=z$, contradicting $y \neq z$. Therefore $s = 1$, which according to (4) means $u=1$. Thus $(x,y) = (d_1,d_1t)$ and $(x,z) = (d_2,d_2v)$, i.e. $x|y$ and $x|z$. q.e.d.

Nice problem! I'll use the notation $\operatorname{lcm}(x,y) := [x,y]$ instead for ease of readability. Observe that $x$ divides $\operatorname{lcm}(x,y)$ and $\operatorname{lcm}(x,z)$, and so $x$ divides $\operatorname{lcm}(x,y) - \operatorname{lcm}(x,z) = y - z$. Thus $y$ and $z$ have the same remainder when divided by $x$, i.e. there exists an integer $a$ such that $y = z + ax$. In turn, the Euclidean Algorithm implies \[ \gcd(x,y) = \gcd(x,z+ax) = \gcd(x,z) =: d, \]and so the original equation rewrites as \begin{align*} y - z = \operatorname{lcm}(x,y) - \operatorname{lcm}(x,z) = \frac{xy}{d} - \frac{xz}d = \frac xd(y-z). \end{align*}Since $y\neq z$, we may divide both sides of the equation by $y-z$ to get $\tfrac xd = 1$. Hence $x = d = \gcd(x,y)$, which implies $x\mid y$, and $x = d = \gcd(x,z)$, which implies $x\mid z$.

Notice that $x\mid\operatorname{lcm}(x,y)-\operatorname{lcm}(x,z)$ so $x\mid y-z.$ Hence, let $y=kx+z$ and we see that $$\gcd(x,y)=\gcd(x,kx+z)=\gcd(x,z).$$Substituting $\operatorname{lcm}(x,y)=xy/\gcd(x,y)$ and $\operatorname{lcm}(x,z)=xz/\gcd(x,z)=xz/\gcd(x,y)$ yields $$\frac{x}{\gcd(x,y)}(y-z)=y-z$$and since $y\neq z$ we conclude $x=\gcd(x,y)$ or $x\mid y.$ Similarly, $x\mid z.$ $\square$