The following expansion is close to your idea and assuming $\log{y}$ is the natural logarithm.
Case 1. Let's assume $x=y$ then $\ln{2y}=\ln^2{y}$. By replacing $\ln{y}=z$ it's easy to check that $P(z)=z^2-z-\ln{2}=0$ has 2 solutions, the largest one $z_2$ is in between $(1,\frac{7}{4})$ and $\forall z > z_2: P(z)>0$. It's easy to check that $z_2 < \frac{7}{4}$ from $\ln{2} < \frac{21}{16}$.
Case 2. Let's assume $x>y$ or $x=qy+r, 0\leq r<y, q>0$ then
$$\ln((q+2)y) > \ln((q+1)y+r)=\ln(qy+r)\ln{y} \geq \ln(qy)\ln{y}$$
or
$$\ln(q+2) + \ln{y} > \ln{q}\ln{y} + \ln^2{y}$$
By replacing $\ln{y}=z$
$$P(z)=z^2+(\ln{q}-1)z-\ln(q+2)<0$$
Case 2.a $q=1$. This means
$$P(z)=z^2-z-\ln{3}<0$$
another easy to check case that the largest solution $z_2$ of $P(z)$ is in between $(1,\frac{7}{4})$ (from $\ln{3} < \frac{21}{16}$) and $\forall z > z_2: P(z)>0$.
Case 2.b $q\geq 2$. We will check the roots of $P(z)$
$$D=\sqrt{(\ln{q}-1)^2+4\ln(q+2)}$$
and
$$\sqrt{(\ln{q}-1)^2+4\ln{q}}<D<\sqrt{(\ln(q+2)-1)^2+4\ln(q+2)}$$
(including for $q=2$) or
$$\ln{q}+1<D<\ln(q+2)+1$$
And the largest of 2 solution for $P(z)$ is
$$z_2=\frac{-(\ln{n}-1)+D}{2}$$
or
$$1<z_2<\ln{\sqrt{1+\frac{2}{q}}}+1<\frac{7}{4}$$
and in fact, with large $q$, is aggressively approaching $1$. In any case, the largest solution $z_2$ is in between $(1,\frac{7}{4})$ and $\forall z > z_2: P(z)>0$.
Note: regarding $$\ln{\sqrt{1+\frac{2}{q}}}+1<\frac{7}{4}$$
For $q\geq 2$ we have $\frac{3}{4} > \frac{1}{2} \ln{2} \geq \ln{\sqrt{1+\frac{2}{q}}} \Rightarrow \frac{7}{4} > \ln{\sqrt{1+\frac{2}{q}}} +1$
In all the cases, we have $\ln{y}=z_2 \in (1,\frac{7}{4})$ or $y \in \{3, 4, 5\}$. The remaining part is to check each $\ln(x+3)=\ln{x}\cdot \ln{3}$, $\ln(x+4)=\ln{x}\cdot \ln{4}$ and $\ln(x+5)=\ln{x}\cdot \ln{5}$ individually, but numerical approaches reveal no integer solutions.