Setting $y = x$ and $y = 1$ yields
$$f(x+1) = 1 + f(x) + \alpha x \qquad (1)$$
and
$$f(x+1) = \frac{f(x)}{f(1)} + f(1) + \alpha x \qquad (2)$$
respectively. Equating (1) and (2) implies
$$f(x)\left(1 - \frac{1}{f(1)}\right) = f(1) - 1.$$
As $f$ cannot be constant due to (1), we obtain $f(1) = 1$. By induction, we get
$$f(x) = \frac{\alpha}{2}x(x-1) + x \quad \text{for all } x \in \mathbb{Z}^+. \qquad (3)$$
In particular, this implies $f(2) = \alpha + 2$ and $f(4) = 6\alpha + 4$. Setting $x = 4$ and $y = 2$ in the functional equation yields
$$\alpha^2 - 2\alpha = 0.$$
Thus we must have $\alpha = 2$ in order to obtain solutions. From now on, we only consider this case.
From (3), we obtain $f(x) = x^2$ for $x \in \mathbb{Z}^+$. By induction, we obtain that for $x \in \mathbb{Q}^+$ and $n \in \mathbb{Z}^+$, from the relation $f(x+n) = (x+n)^2$ it follows that $f(x) = x^2$.
Let now $\frac{a}{b} \in \mathbb{Q}^+$ with $a, b \in \mathbb{Z}^+$. We set $x = a$ and $y = b$ and obtain
$$f\left(\frac{a}{b} + b\right) = \frac{a^2}{b^2} + b^2 + 2a = \left(\frac{a}{b} + b\right)^2.$$
The above remark implies that $f(\frac{a}{b}) = (\frac{a}{b})^2$. It is easily verified that $f(x) = x^2$ is indeed a solution.
Thus there is no solution for $\alpha \neq 2$ and the solution $f(x) = x^2$ for $\alpha = 2$.