We proceed by induction. The case $n=1$ is clear, since we may pick $k_0 = 3$ and $k_1 = 7$. Let us assume now that for a certain $n$ there are pairwise relatively prime integers $1 < k_0 < k_1 < \cdots < k_n$ such that $k_0 k_1 \cdots k_n - 1 = a_n(a_n - 1)$, for some positive integer $a_n$. Then choosing $k_{n+1} = a_n^2 + a_n + 1$ yields
$$k_0 k_1 \cdots k_{n+1} = (a_n^2 - a_n + 1)(a_n^2 + a_n + 1) = a_n^4 + a_n^2 + 1,$$
so $k_0 k_1 \cdots k_{n+1} - 1$ is the product of the two consecutive integers $a_n^2$ and $a_n^2 + 1$. Moreover,
$$\gcd(k_0 k_1 \cdots k_n, k_{n+1}) = \gcd(a_n^2 - a_n + 1, a_n^2 + a_n + 1) = 1,$$
hence $k_0, k_1, \dots, k_{n+1}$ are pairwise relatively prime. This completes the proof.

Write the relation to be proved as
$$4k_0 k_1 \cdots k_n = 4a(a+1) + 4 = (2a+1)^2 + 3.$$
There are infinitely many primes for which $-3$ is a quadratic residue. Let $2 < p_0 < p_1 < \dots < p_n$ be such primes. Using the Chinese Remainder Theorem to specify $a$ modulo $p_n$, we can find an integer $a$ such that $(2a+1)^2 + 3 = 4p_0 p_1 \cdots p_n m$ for some positive integer $m$. Grouping the factors of $m$ appropriately with the $p_i$'s, we obtain $(2a+1)^2 + 3 = 4k_0 k_1 \cdots k_n$ with $k_i$ pairwise relatively prime. We then have $k_0 k_1 \cdots k_n - 1 = a(a+1)$, as desired.

We are supposed to show that for every positive integer $n$, there is a positive integer $x$ such that $x(x+1)+1 = x^2+x+1$ has at least $n$ distinct prime divisors. We can actually prove a more general statement.

Claim. Let $P(x) = a_d x^d + \cdots + a_1 x + 1$ be a polynomial of degree $d \ge 1$ with integer coefficients. Then for every positive integer $n$, there is a positive integer $x$ such that $P(x)$ has at least $n$ distinct prime divisors.

The proof follows from the following two lemmas.

Lemma 1. The following set is infinite:
$$Q = \{p \mid p \text{ a prime for which there is an integer } x \text{ such that } p \text{ divides } P(x)\}.$$
Proof. The proof is analogous to Euclid's proof that there are infinitely many primes. Namely, if we assume that there are only finitely many primes $p_1, p_2, \dots, p_k$ in $Q$, then for each integer $m$, $P(m p_1 p_2 \cdots p_k)$ is an integer with no prime factors, which must equal 1 or $-1$. However, since $P$ has degree $d \ge 1$, it takes each of the values 1 and $-1$ at most $d$ times, a contradiction. $\square$

Lemma 2. Let $p_1, p_2, \dots, p_n$, $n \ge 1$ be primes in $Q$. Then there is a positive integer $x$ such that $P(x)$ is divisible by $p_1 p_2 \cdots p_n$.
Proof. For $i = 1, 2, \dots, n$, since $p_i \in Q$ we can find an integer $c_i$ such that $P(x)$ is divisible by $p_i$ whenever $x \equiv c_i(\text{mod } p_i)$. By the Chinese Remainder Theorem, the system of $n$ congruences $x \equiv c_i(\text{mod } p_i)$, $i = 1, 2, \dots, n$ has positive integer solutions. For every positive integer $x$ that solves this system, $P(x)$ is divisible by $p_1 p_2 \cdots p_n$. $\square$