When $\alpha n + \beta \sqrt{n}$ is an integer for infinitely many n's?

Yes, $\alpha$ should always be rational. To prove this, suppose that we have $$\alpha n+\beta \sqrt{n}=a$$ and $$\alpha m+\beta\sqrt{m}=b$$ for some integers $a$ and $b$. Solving this system of linear equations in $\alpha$ and $\beta$ we get

$$\beta=\frac{am-bn}{\sqrt{nm}(\sqrt{n}-\sqrt{m})}=\frac{am-bn}{n-m}\left(\frac{1}{\sqrt{m}}+\frac{1}{\sqrt{n}}\right)$$

Therefore, as $\beta\neq 0$ (as otherwise we have trivally $\alpha\in \mathbb Q$), we have for any pair $(n_1,n_2)$ with $\alpha n_i+\beta \sqrt{n_i}\in \mathbb Z$ that $\mathbb Q(\beta)=\mathbb Q(1/\sqrt{n_1}+1/\sqrt{n_2})$. Therefore, $\mathbb Q(\beta)$ is (at most) biquadratic extension. Therefore, if squarefree parts of $n_1$ and $n_2$ are different, we have $\mathbb Q(\beta)=\mathbb Q(\sqrt{n_1}, \sqrt{n_2})$. The latter field has only three quadratic subextensions, so there are only four possibilities for the value of square-free part of $n$ if $\alpha n+\beta\sqrt{n}$ is integer. Thus, there exist $n,m$ with equal squarefree parts and with this property, in which case we have $\mathbb Q(\beta)=\mathbb Q(\sqrt n)$ and

$$\alpha n=\frac{a\sqrt{nm}+bn}{\sqrt{nm}-m}$$

for some integers $a$ and $b$ (this follows by solving the system mentioned before). As the squarefree part of $n$ and $m$ coincide, we have

$$\alpha\in \mathbb Q$$

and

$$\beta\in \sqrt{n}\mathbb Q,$$

which proves both of your assertions.