Your argument is correct. There was a verbal slip, in that it is not true that you simplified $\frac{\log n}{n^2}$ to $\frac{\log(n+1)}{\log n}\cdot \frac{n^2}{(n+1)^2}$. What you did was to simplify something else, which is a ratio of successive coefficients.
Applying the Squeeze Theorem, though well done, was unnecessary. You are trying to find $\lim_{n\to\infty}\frac{\log(n+1)}{\log n}\cdot \frac{n^2}{(n+1)^2}.$ There is a general result that says that if $\lim_{n\to \infty}a_n=a$ and $\lim_{n\to \infty}b_n=b$, where $a$ and $b$ are real numbers, then $\lim_{n\to \infty}a_nb_n=ab$ (the limit of a product is the product of the limits). That's all you need.
Comment: Getting practice in using the Squeeze Theorem is not a bad idea. However, inequalities of the type you found are not always so easy to get. That's why certain basic tools, about the limit of a sum, and the limit of a product, are so useful. Another very useful result if that if the sequence $(a_n)$ has limit $a$, where $a$ is a real number, and $f$ is a continuous function, then the sequence $(f(a_n))$ has limit $f(a)$.