We shall prove a more general statement in the following lines.
Proposition. Let $k$ be a positive integer. Prove that for all pairs $(c, d)$ of integers such that $|c|$ and $|d|$ are prime numbers there is a positive integer $N$ such that the number of pairs $(a, b)$ of integers satisfying the following inequalities is at most $N$, for all sufficiently small positive real numbers $u, v$.
$$|b||a-c|^{2021} \le u|c|^{1011}, \quad |a^k b - c^k d| \le v|c|^{k-1010}.$$
Let us rewrite the original inequalities as $|b|\left|\frac{a}{c} - 1\right|^{2021} \le u|c|^{-1010}$ and $\left|\left(\frac{a}{c}\right)^k b - d\right| \le v|c|^{-1010}$. Then if $k \le 1010$ then $a^k b = c^k d$ and hence, we only have finitely many solutions for $a, b$. Hence, assume that $k > 1010$. We then prove the following lemma:
Lemma. There are polynomials $P(x), Q(x)$ of degree $1010$ with integer coefficients such that the polynomial $P(x) - x^k Q(x)$ has a $2021$-fold zero at $x = 1$.
Proof. We shall prove that for all triples $(m, n, k)$ of natural numbers such that $n > k$ there are polynomials $P(x) = P_{(m,n,k)}(x)$, $Q(x) = Q_{(m,n,k)}(x)$ of degrees $m, n$ with rational coefficients such that $P(x) - x^k Q(x)$ is divisible by $(x-1)^{m+n+1}$. We shall prove this through induction, considering the following expression:
$$x (P_{(m,n-1,k)}(x) - x^k Q_{(m,n-1,k)}(x)) + r (P_{(m-1,n,k)}(x) - x^k Q_{(m-1,n,k)}(x)).$$
Since both brackets are divisible by $(x-1)^{m+n}$ then there would be a rational number $r$ such that the above expression is divisible by $(x-1)^{m+n+1}$ then defining $P_{(m,n,k)}(x) = xP_{(m,n-1,k)}(x) + rP_{(m-1,n,k)}(x)$ and $Q_{(m,n,k)}(x) = xQ_{(m,n-1,k)}(x) + Q_{(m-1,n,k)}(x)$. Moreover, for the base case, that is when one of $m, n$ are zero, the statement is obvious. This completes our proof.

Comment 1. If $P_{(m,n,k)}(0) = 0$ for some non-zero $m, n$ then the polynomial $P_{(m,n,k)}(x) - x^k Q_{(m,n,k)}(x)$ has at most $m+n$ non-zero coefficients, impossible.
Lemma. $P_{(m,n,k)}(x), Q_{(m,n,k)}(x)$ have no common factor.
Proof. Assume for the contradiction that there is a non constant polynomial $D(x)$ of degree $d$ dividing both of them. Then at least $(x-1)^{m+n+1-d}$ divides $\frac{P_{(m,n,k)}(x)}{D(x)} - x^k \frac{Q_{(m,n,k)}(x)}{D(x)}$. Since the not-zero coefficient of this polynomial is at most $m+n-2d+2$. But, according to our lemma 2 it must at least have $m+n+2-d$ non-zero coefficients. Therefore, $d \le 0$. This completes our proof.

Now, there are coprime polynomials $P(x), Q(x)$ of degree $1010$ with integer coefficients such that $P(x) - x^k Q(x)$ is divisible by $(x-1)^{2021}$ and hence there are polynomials $A(x), B(x)$ with integer coefficients of the same degree that doesn't exceed $1010$ such that
$$A(x)P(x) + B(x)Q(x) = R, \quad (1)$$
for some non-zero constant $R$.
Lemma. If $a, b, c$ and $d$ satisfy the statement of our problem then $bP\left(\frac{a}{c}\right) = dQ\left(\frac{a}{c}\right)$.
Proof. We prove the lemma for polynomial $P(x) - x^k Q(x)$, for some positive integer $k \ge \deg P(x) = \deg Q(x) = d$. Then, the second inequality becomes $|a^k b - c^k d| \le v|c|^{k-1010}$. Since $|b| \ge 1$, take $u < 1$ then $|a-c| \le |c|$ therefore, $|a| \le 2|c|$. Moreover, let $d = 1010$, and $R(x) = P(x) - x^k Q(x)$, the following quantity is an integer:
$$c^d \left( bP\left(\frac{a}{c}\right) - dQ\left(\frac{a}{c}\right) \right) = bc^d R\left(\frac{a}{c}\right) + (a^k b - c^k d) c^{d-k} Q\left(\frac{a}{c}\right).$$
Now, if $bP(\frac{a}{c}) \neq dQ(\frac{a}{c})$ then, $|c^d (bP(\frac{a}{c}) - dQ(\frac{a}{c}))| \ge 1$. Thus, by triangle inequality, we find that
$$|bc^d R\left(\frac{a}{c}\right)| + \left| (a^k b - c^k d) c^{d-k} Q\left(\frac{a}{c}\right) \right| \ge 1.$$
Write $R(x) = (1-x)^{2d+1} S(x)$ then, $bc^d R\left(\frac{a}{c}\right) = bc^d (1-\frac{a}{c})^{2d+1} S\left(\frac{a}{c}\right) = c^{-d-1} b(c-a)^{2d+1} S\left(\frac{a}{c}\right)$. Thus,
$$|c^{-d-1} b(c-a)^{2d+1} S\left(\frac{a}{c}\right)| + \left| (a^k b - c^k d) c^{d-k} Q\left(\frac{a}{c}\right) \right| \ge 1.$$
That is, $|b||c-a|^{2d+1} |c^{-d-1}| |S(\frac{a}{c})| + |a^k b - c^k d| |c^{d-k}| |Q(\frac{a}{c})| \ge 1$. By use of problem's assumption, we find for all $u, v$: $|u| |S(\frac{a}{c})| + |v| |Q(\frac{a}{c})| \ge 1$.

Assume that $S(x) = c_{k-d-1}x^{k-d-1} + \cdots + c_0$ be a polynomial with integer coefficients. Then, $|S(\frac{a}{c})| \le \sum_{i=0}^{k-d-1} |c_i|2^i = C_1$ which is a constant number. Furthermore, by the same argument we can find that $|Q(\frac{a}{c})| \le C_2$ for some constant $C_2$. This yields to:
$$u C_1 + vC_2 \ge 1.$$
The last inequality would be absurd for all sufficiently small $u, v$. We are done.

Putting $x = \frac{a}{c}$ in (1) and multiply both sides by $c^{2020}$ it follows that
$$c^{1010}A\left(\frac{a}{c}\right) c^{1010}P\left(\frac{a}{c}\right) + c^{1010}B\left(\frac{a}{c}\right) c^{1010}Q\left(\frac{a}{c}\right) = c^{2020}R.$$
Now, using lemma 4 yielding to
$$c^{1010} P\left(\frac{a}{c}\right) \left( d c^{1010} A\left(\frac{a}{c}\right)+b c^{1010} B\left(\frac{a}{c}\right)\right) = c^{2020} d R.$$
Note that if $\gcd(a, c) > 1$ then $c$ must divide $a$. But then, $|b||a-c|^{2021} \le u|c|^{1011}$ becomes $|b|^{\frac{a}{c}-1}|c|^{2021} \le u|c|^{-1010}$. Hence, it follows that $a=c$. Thus, assume that $\gcd(a, c) = 1$.
Hence, $c^{1010}P(\frac{a}{c})$ divides $c^{2020}dR$. Note that
$$\gcd\left(c^{1010}P\left(\frac{a}{c}\right), c\right) = \gcd\left(c, a_{1010}a^{1010}\right) = \gcd(c, a_{1010}) = g \le |a_{1010}|,$$
where $a_{1010}$ is the leading coefficient of $P(x)$. Writing $c = gf$, $\gcd(f, a_{1010}) = 1$ it follows that $c^{1010}P(\frac{a}{c})$ divides $g^{2020}dR$. Since the number of divisors of $Rdg^{2020}$ is at most $2|R|g^{2020} \le 2|R|a_{1010}^{2020}$, hence the total number of possibilities for $a$ is at most $2020|R|a_{1010}^{2020}$. Finally for each certain $a, b = \frac{dQ(\frac{a}{c})}{P(\frac{a}{c})}$ would be determined uniquely. Moreover, note that $P(\frac{a}{c}) \ne 0$ because otherwise, $P, Q$ will have a common factor. Since $R$ only depends on the coefficients of $P(x), Q(x)$ we can put $N = 2020|R|a_{1010}^{2020}$. ■