Let $x$ and $y$ be two positive integers such that
$$\operatorname{lcm}\left(x+2{,}y+2\right)-\operatorname{lcm}\left(x+1{,}y+1\right)=\operatorname{lcm}\left(x+1{,}y+1\right)-\operatorname{lcm}\left(x{,}y\right)$$
Prove that one of the two numbers $x$ and $y$ divide the other.
So far I have been able to prove only this:
$$\operatorname{lcm}\left(x+2{,}y+2\right), \ \operatorname{lcm}\left(x+1{,}y+1\right), \ \operatorname{lcm}\left(x{,}y\right)$$ are in A.P.
I tried assuming the middle term as $m$ and the common difference as $d$ but nothing came after this. Any help will be greatly appreciated.
Unfortunately, I don't see any way to directly use that the $3$ $\operatorname{lcm}$ values are in an arithmetic progression. Instead, the problem can be solved by using that the $\operatorname{lcm}$ values are bounded integral multiples of both of its integers, in particular the larger ones.
Rearranging the given equation gives
$$2\operatorname{lcm}(x + 1, y + 1) = \operatorname{lcm}(x, y) + \operatorname{lcm}(x + 2, y + 2) \tag{1}\label{eq1A}$$
If $x = y$, then \eqref{eq1A} is true and also $x \mid y$. Otherwise, using the symmetry of $x$ and $y$, WLOG let $y \gt x$, so $x \le y - 1$. As asked & answered in several posts here, e.g., at Proving gcd($a,b$)lcm($a,b$) = $|ab|$, we get that
$$\operatorname{lcm}(x, y) = ay, \; \; 1 \le \left(a = \frac{x}{\gcd(x, y)}\right) \le y - 1 \tag{2}\label{eq2A}$$
$$\operatorname{lcm}(x + 1, y + 1) = b(y + 1), \; \; 1 \le \left(b = \frac{x + 1}{\gcd(x + 1, y + 1)}\right) \le y \tag{3}\label{eq3A}$$
$$\operatorname{lcm}(x + 2, y + 2) = c(y + 2), \; \; 1 \le \left(c = \frac{x + 2}{\gcd(x + 2, y + 2)}\right) \le y + 1 \tag{4}\label{eq4A}$$
Substituting the left sides of \eqref{eq2A}, \eqref{eq3A} and \eqref{eq4A} into \eqref{eq1A} results in
$$\begin{equation}\begin{aligned} 2b(y + 1) & = ay + c(y + 2) \\ 2by + 2b & = ay + cy + 2c \\ 2by - ay - cy & = 2c - 2b \\ (2b - a - c)y & = 2(c - b) \end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Using the $\min$ and $\max$ values of the right sides of \eqref{eq3A} and \eqref{eq4A} gives
$$\begin{equation}\begin{aligned} 2(1 - y) & \le 2(c - b) \le 2((y + 1) - 1) \\ -2y \lt -2y + 2 & \le 2(c - b) \le 2y \end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Since \eqref{eq5A} shows that $y \mid 2(c - b)$, then the limits in \eqref{eq6A} allows only $4$ possibilities:
$2(c - b) = -y$ and $2b - a - c = -1$. Adding these together gives $c - a = -y - 1$. However, using \eqref{eq2A} and \eqref{eq4A} gives that $c - a \ge 1 - (y - 1) = -y + 2$, so this case is not possible.
$2(c - b) = 0$ and $2b - a - c = 0$. Using $c = b$ from the first equation in the second one gives $b - a = 0 \; \to \; b = a$. From \eqref{eq2A} and \eqref{eq3A} we have $a \mid x$ and $b \mid x + 1$. Since $a = b$, then $a \mid x + 1$. Thus, $a \mid \gcd(x, x + 1) = 1$, so $a = b = c = 1$. Since the problem's request is equivalent to showing $a = 1$ in \eqref{eq2A}, this case is both valid and solves the problem.
$2(c - b) = y$ and $2b - a - c = 1$. Adding these together gives $c - a = y + 1$. However, using \eqref{eq2A} and \eqref{eq4A} gives that $c - a \le (y + 1) - 1 = y$, so this case is not possible.
$2(c - b) = 2y$ and $2b - a - c = 2$. Adding these together gives $c - a = 2y + 2$ which, as shown in case $3$ above, is also not valid.
Only case $2$ is possible, with this not only showing that $x \mid y$, as requested, but also that $x + 1 \mid y + 1$ and $x + 2 \mid y + 2$. An example solution is $x = 1$ and $y = 7$.