We will prove that the number $2^{2022}7^k$ is not sureño and has exactly 2022 sureño divisors (for every positive integer $k$).

Every power of 2 is sureño. Indeed, the divisors of $2^k$ are $2^j$ for $j = 0, \dots, k$ and $d_j - d_{j-1} = 2^j - 2^{j-1} = 2^{j-1}$ is a divisor of $2^k$. Therefore we have that the powers of 2 that divide $2^{2022} \cdot 7^k$ are sureño, that is, $2, 2^2, 2^3, \dots, 2^{2022}$. We have showed that the number has 2022 sureño divisors.

It remains for us to prove that the other divisors as well as the number itself are not sureño. That is, we have to show that the numbers $2^j \cdot 7^k$ are not sureño for $j \ge 0$ and $k \ge 1$.

If $j = 0$ then the first divisors of the number are 1 and 7. Hence the number is not sureño, since $7 - 1 = 6$ is not a divisor of $2^j \cdot 7^k$.

If $j = 1$ then the first divisors of the number are 1, 2 and 7. Again, the number is not sureño, since $7 - 2 = 5$ is not a divisor of $2^j \cdot 7^k$.

Finally, if $j \ge 2$ then the first divisors of the number are 1, 2, 4 and 7. Hence the number is not sureño, since $7 - 4 = 3$ is not a divisor of $2^j \cdot 7^k$.

We have completed the proof.