Let $D_n$ be a discriminant of quadratic equation $x^2 + p_n x + q_n = 0$, for each $n \in \mathbb{N}$, i.e.
$$D_n = p_n^2 - 4q_n.$$
By assumption we conclude that $D_1$ is a square of an integer.
Further, we have:
$$D_{n+1} = p_{n+1}^2 - 4q_{n+1} = (p_n + 1)^2 - 4 \left( q_n + \frac{1}{2}p_n \right) = p_n^2 - 4q_n + 1 = D_n + 1.$$
Let's suppose that the equation $x^2 + p_n x + q_n = 0$ has two integer solutions for some $n$. Then $D_n = k^2$ for some integer $k$, hence, we have:
$$D_{n+2k+1} = D_n + 2k + 1 = k^2 + 2k + 1 = (k+1)^2.$$
Moreover, since $\frac{1}{2}(-p_n + \sqrt{D_n})$ and $\frac{1}{2}(-p_n - \sqrt{D_n})$ are integers, we conclude that $p_n$ and $D_n$ are of the same parity. But,
$$p_{n+2k+1} \equiv p_n + 2k + 1 \equiv p_n + 1 \equiv D_n + 1 \equiv D_n + 2k + 1 \equiv D_{n+2k+1} \pmod{2}$$
so the equation $x^2 + p_{n+2k+1}x + q_{n+2k+1} = 0$ also has two integer solutions.
Hence, we proved that there are infinitely many integers $n$ for which the equation $x^2 + p_n x + q_n = 0$ has two integer solutions.