There exists n such that number of primes between $(n+1)^{2021}$ and $n^{2021}$ is greater than 1122021140.

Let $\epsilon$, $M$ be positive numbers. Then there exists $n$ such that between $n^{1+\epsilon}$ and $(n+1)^{1+\epsilon}$ there are more than $M$ prime numbers. Indeed, otherwise a simple estimate would show that $$\sum_{p \textrm{ prime}} \frac{1}{p} \le M \cdot \sum_{n=1}^{\infty} \frac{1}{n^{1+\epsilon}}< \infty$$ contradiction

$\bf{Added:}$ A similar argument shows that for every increasing sequence $a_n$ with $\sum \frac{1}{a_n} < \infty$ we have $$\lim \sup_{n\to \infty} (\pi(a_{n+1}) -\pi(a_n)) = +\infty$$

More directly, $n=1$ will work. There are many more than $1122021140$ primes between $1$ and $2^{2021}$, and we can prove it.

For example, there is a lower bound on $\pi(2n)$ via the prime factorization of $\binom{2n}{n}$ (for details, see this MSE post) that tells us $$ \pi(2n)\ge \frac{n}{\log_2 (2n)} $$ so in particular, $\pi(2^{2021}) \ge \frac{2^{2020}}{2021}$. This is much larger than $1122021140$: it's a number with over $600$ digits.

From the question you link:

There are unconditional results by Huxley and Heath-Brown showing $(1)$ for $h=y-x$ roughly being $x^{7/12}$.

Since $h\in O(n^{2020})=O(x^{2020/2021})=\omega(x^{7/12})$, there is an unconditional proof of $(1)$. Then $(2)$ follows by simply comparing growth rates of numerator and denominator.