Consider the line $y = ax + b$ where $b = 2$ and $a = p_1 p_2 \cdots p_m$ for primes $p_1, p_2, \ldots, p_m$ that will be chosen appropriately. We need to ensure that for a rational point $(x, y)$, $xy = z \in \mathbb{Z}$ such that $1 + az$ is a perfect square.

We construct the primes $p_1, p_2, \ldots, p_m$ such that $p_i > 2017^{2017}$ and for all $1 \le j \le m-1$,
$$3 \prod_{k \ne j, 1 \le k \le m} p_k \equiv 2 \pmod{p_j}.$$
This can be achieved by ensuring $p_m \equiv \frac{2}{3 \prod_{k \ne j, 1 \le k \le m-1} p_k} \pmod{p_j}$, which is guaranteed by the Chinese Remainder Theorem and Dirichlet's theorem.

We claim that this construction works. Suppose $1 + az \equiv x^2 \pmod{a}$. Then for some $x_1, x_2, \ldots, x_m \in \{-1, 1\}$, $x \equiv x_j \pmod{p_j}$.

Let $v_j$ be the unique integer such that $v_j \equiv 0 \pmod{p_i}$ for all $i \ne j$ and $v_j \equiv 2 \pmod{p_j}$ with $1 \le v_j \le P$. This implies that the set of $x$ such that $x^2 \equiv 1 \pmod{a}$ in $\mathbb{Z}_a$ is of the form $-1 + \sum_{j=1}^m e_j v_j$ where $e_j \in \{0, 1\}$. Notice $v_j = \frac{3a}{p_j}$ for $1 \le j \le m-1$. For size reasons, $2 < v_1 + \cdots + v_{m-1} = 3a \sum_{j=1}^{m-1} \frac{1}{p_j} < a$. Therefore, $2 < v_1 + \cdots + v_m < 2a$. Since $v_1 + \cdots + v_m \equiv 2 \pmod{p_j}$ for all $1 \le j \le m$, it follows that $v_1 + \cdots + v_m = a + 2$.

Step 1: Construct an interval with $2017 + 58 = 2075$ blue points and 0 red points. Observe that the set $\sum_{j=1}^{m-1} e_j v_j \equiv 3 \sum e_j \pmod{6}$. Therefore, if $x = -1 + \sum_{j=1}^{m-1} e_j v_j$, $3 \mid x^2 - 1$ (so $3 \mid \frac{x^2 - 1}{a}$) and the parity of $\frac{x^2 - 1}{a}$ is also the same as the parity of $x^2 - 1$, which is the parity of $\sum e_j$.

Therefore, all $z$ such that $1 + az = (\sum_{j=1}^{m-1} e_j v_j - 1)^2$ for some $e_1 + \cdots + e_{m-1}$ odd corresponds to a blue point because $3 \mid z$ and $2 \nmid z$ and $\frac{a}{4} x^2 + x = z$ has a solution with $x \in \mathbb{Q}$. Hence, when $0 < z < (\sum_{j=1}^{m-1} v_j - 1)^2$, there is an interval of $2^{m-2} > 2075$ blue points.

Step 2: Use discrete continuity.

Suppose we sort all $z_1 < z_2 < \cdots < z_{6 \times 2^m}$ such that $1 + az_i = b_i^2$ is a perfect square for all $i$. Then notice $b_{i + 2^m} = b_i + a$ because there are $2^m$ solutions to $x^2 \equiv 1 \pmod{a}$ in $\mathbb{Z}_a$.

Consider $z_{j + t2^m}$ for $0 \le t \le 5, 1 \le j \le 2^m$. We can see $z_{j + t2^m} = \frac{b_{j + t2^m}^2 - 1}{a} = \frac{(b_j + ta)^2 - 1}{a}$.

We know $a$ is either $1$ or $-1$ mod 6. For any value of $b_j$, we can set $t \in \{0, \cdots, 5\}$ such that $3 \mid b_j + ta$ and $2 \nmid b_j + ta$, which forces $\frac{(b_j + ta)^2 - 1}{a} = z_j$ to be divisible by 2 but not 3, which is red. Therefore, the number of red points among $z_1 < \cdots < z_{6 \times 2^m}$ is at least $2^m$, while the number of blue points is at most $5 \times 2^m$.

Let $c_j$ be the number of blue points among $z_j, \cdots, z_{j + 2074}$. Observe $|c_{j+1} - c_j| \le 1$ and $c_t = 2075$ for some $t$. Therefore, by discrete continuity, there exists $c_j = 2017$, finishing the problem.

The answer is: $\boxed{\text{Yes}}.$