Lemma 1. If a positive integer $n$ satisfies $24|n+1$ then the sum of its positive divisors $\sigma(n)$ is divisible by 24.

Proof. Indeed, if $d$ is a divisor of $n$ then $\frac{n}{d}$ is also a divisor of $n$. Because $n \equiv 2 \pmod{3}$ so it cannot be a perfect square, which means the sum of its divisors can be divided into pairs of the form
$$d + \frac{n}{d} = \frac{d^2 + n}{d}.$$
Note that $n \equiv 2 \pmod{3}$ and $d^2 \equiv 1 \pmod{3}$ so the sum above is divisible by 3.
On the other hand, $n \equiv 7 \pmod{3}$ and $d \equiv 1, 3, 5, 7 \pmod{8}$, which implies $d^2 \equiv 1 \pmod{8}$ so the above sum is also divisible by 8. Since $(3, 8) = 1$ then the above sum is divisible by 24 and the lemma is proved. $\square$

a.
Back to our problem, denote $y_n = x_n x_{n+1} + x_{n+1} x_{n+2} + x_{n+2} x_{n+3}$, we need to prove that $\sigma(y_n + 2018)$ is divisible by 24. We also have $2018 \equiv 2 \pmod{24}$ so according to the lemma, we need to show $y_n \equiv -3 \pmod{24}$.

Consider the period of the remainder when being divided by 3 of the sequence $(x_n)$, note that $x_{n+2} \equiv x_{n+1} - x_n + 1 \pmod{3}$ we have 2, 0, 2, 0, 2, 0, ... this sequence is periodic with period 2 and
$$y_n \equiv 0 \cdot 2 + 2 \cdot 0 + 0 \cdot 2 = 0 \pmod{3}.$$

Similarly, consider the remainder of $(x_n)$ when being divided by 8, note that $x_{n+2} \equiv -x_{n+1} - x_n \pmod{8}$ we have
$$2, 3, 3, 2, 3, 3, 2, 3, 3, \dots$$
which means this sequence is periodic with period 3 and
$$y_n \equiv 2 \cdot 3 + 3 \cdot 3 + 3 \cdot 2 = 5 \pmod{8}.$$
It follows that $(y_n + 3)$ is both divisible by 3, and 8, so $y_n \equiv -3 \pmod{24}$ and a) is proved.

b.
Now, we prove the following lemma.

Lemma 2. Consider the integer sequence $(z_n)$ satisfying $z_{n+2} = a z_{n+1} - z_n + b$ then the following quantity is constant
$$z_{n+1}^2 - z_n z_{n+2} - b z_{n+1} \text{ for all } n \ge 0.$$
Proof. Indeed, we have the following transformation
$$\begin{aligned} z_{n+1}^2 - z_n z_{n+2} - b z_{n+1} &= z_{n+1}(z_{n+1} - b) - z_n(a z_{n+1} - z_n + b) \\ &= z_{n+1}(a z_n - z_{n-1}) - z_n(a z_{n+1} - z_n + b) \\ &= z_n^2 - z_{n-1} z_{n+1} - b z_n. \end{aligned}$$
The above equality holds for all $n \ge 0$ so $z_{n+1}^2 - z_n z_{n+2} - b z_{n+1} = z_1^2 - z_0 z_2 - b z_1 = c$ where $c$ is a constant. $\square$

Thus, there exists $C \in \mathbb{Z}$ such that $x_{n+1}^2 - x_n x_{n+2} - 280 x_{n+1} = C$. We have
$$\begin{aligned} x_{n+1}^2 - x_n(7 x_{n+1} - x_n + 280) - 280 x_{n+1} &= C \\ x_{n+1}^2 + x_n^2 - 7 x_{n+1} x_n - 280(x_{n+1} + x_n) &= C \\ (x_{n+1} + x_n - 140)^2 &= 9(x_{n+1} x_n + 2019) - 9 \cdot 2019 + C + 140^2 \\ u_n^2 &= v_n^2 + C + 1429, \end{aligned}$$
where $u_n = x_{n+1} + x_n - 140$, $v_n = 3\sqrt{x_{n+1} x_n + 2019}$ for all $n \ge 0$.

Since $(x_n)$ is an increasing integer sequence so it is unbounded, thus $(u_n)$ is increasing and unbounded. It is also clear that if $x_n x_{n+1} + 2019$ is a perfect square then $v_n \in \mathbb{Z}^+$.

Hence, $u_n + v_n | C + 1429$ for infinite values of $n$. Clearly, this case only happens when $C + 1429 = 0$ so
$$(x_{n+1} + x_n - 140)^2 = 9(x_{n+1} x_n + 2019), \forall n \in \mathbb{Z}^+.$$
We have $(x_0 + x_1 - 140)^2 \ge 2019 \cdot 9 > 44^2 \cdot 3^2 = 132^2$ so $|140 - x_0 - x_1| \ge 133$, but $0 \le x_0 < x_1 < 101$ then $140 - (x_0 + x_1) \ge 133$, i.e. $x_0 + x_1 \le 7$. We also have
$$C = x_1^2 + x_0^2 - 7 x_1 x_0 - 280(x_1 + x_0) = -1429.$$
Notice that $x_1^2 + x_0^2 \le 49$ so $-1429 = C < 49 - 280(x_1 + x_0)$, which implies $x_0 + x_1 \ge 5$. By direct checking, the case $x_1 + x_0 = 7$ and $x_1 + x_0 = 6$ has no solution. So $x_0 + x_1 = 5$, which implies $x_0 x_1 = 6$ so it's easy to see that $x_0 = 2, x_1 = 3$.

Therefore, $(x_0, x_1) = (2, 3)$ is the only satisfying pair. $\square$