$$\ln^2(1+x)\sim x^2-x^3,x\rightarrow \infty\Rightarrow \int_1^{+\infty} \frac{\ln^2(1+x)}{x^{2\alpha}}\mathrm dx=\int_1^{+\infty} \frac{x^2-x^3}{x^{2\alpha}}\mathrm dx=$$ $$\int_1^{+\infty} \frac{1}{x^{\alpha}}\mathrm dx-\int_1^{+\infty} x^{1-\alpha}\mathrm dx=\lim\limits_{t\to\infty}\int_1^{t} \frac{1}{x^{\alpha}}\mathrm dx-\lim\limits_{t\to\infty}\int_1^{+\infty} x^{1-\alpha}\mathrm dx$$
First limit exists if $\alpha>1$ and second if $\alpha>2$.
Integral is convergent if $\alpha>2$ and divergent if $\alpha\in (-\infty,1]\cup[1,2]$.
Is this correct?
No, this is not correct. The statement
$$\ln^2(1 + x) \sim x^2 - x^3, \quad x \to \infty$$
is not valid: Note that the right side is negative for $x > 1$ while the left side is not. It's also a massive overestimate of the logarithm: Since $\ln x \le x^{\epsilon}$ for all $\epsilon > 0$ and sufficiently large $x$, estimating $\ln$ by $x^3$ is way too large. In fact, the logarithm is almost negligible.
There's also a problem with your algebra: You move from $x^{2} / x^{2\alpha}$ and simplify it as $1/x^{\alpha}$, which is not correct.
For a different approach, use the bound I mentioned above. For example, if $x > 1000$ (where $1000$ is really 'some big number') then $x^{1/10} > \ln^2(1 + x)$. This gives the estimate
$$\int_{1000}^{\infty} \frac{\ln^2(1 + x)}{x^{2\alpha}} dx \le \int_{1000}^{\infty} \frac{x^{1/10}}{x^{2\alpha}} = \int_{1000}^{\infty} x^{\frac 1 {10} - 2\alpha} dx$$
This converges if $\frac 1 {10} - 2\alpha < -1$, or $\alpha > 5/11$. Use a similar idea with different exponents to prove that the integral converges for all $\alpha > 1/2$.
On the other hand, if $\alpha \le \frac 1 2$, the integral can be estimated from below by simply ignoring the logarithm:
$$\int_1^{\infty} \frac{\ln^2(1 + x)}{x^{2\alpha}} \, dx \ge \int_1^{\infty} \frac{1}{x^{2\alpha}} \, dx = \infty$$ since the exponent is too small.