Answer: $(756; 1008)$, $(600; 960)$.

Let $d$ denote the greatest common divisor of $a$ and $b$, i.e., $a = da_1$, $b = db_1$, where $\gcd(a_1, b_1) = 1$. Then the given equality can be presented in the form
$$d(a_1^3 + b_1^3) = 1911a_1b_1.$$
It follows that $d a_1^3 : b_1$, and, since $a_1$ and $b_1$ are coprime, $d : b_1$. Similarly, $d : a_1$. Hence, $d : a_1b_1$ because $a_1$ and $b_1$ are coprime. Let $d = d_1 a_1 b_1$, then the equation can be presented in the form
$$d_1(a_1^3 + b_1^3) = 1911, \quad a_1 \le b_1. \qquad (1)$$
Note that $a_1^3 + b_1^3 \ge (a_1 + b_1)^3/4$, whence $4 \cdot 1911 \ge d_1 \cdot (a_1 + b_1)^3$; therefore, $8000 > (a_1 + b_1)^3$, or $a_1 + b_1 < 20$.

Further, $a_1^3 + b_1^3 = (a_1 + b_1)^3 - 3 a_1 b_1 (a_1 + b_1)$, and (1) can be rewritten as
$$d_1(a_1 + b_1)((a_1 + b_1)^2 - 3 a_1 b_1) = 1911. \qquad (2)$$
We see that $a_1 + b_1$ is a divisor (less than 20) of $1911 = 3 \cdot 7^2 \cdot 13$. Hence exactly three cases are possible.

1. $a_1 + b_1 = 3$, $a_1 = 1$, $b_1 = 2$. Then $d_1 \cdot 3 \cdot (9 - 6) = 1911$, but $9$ is not a divisor of $1911$.

2. $a_1 + b_1 = 7$. If $a_1 = 1$, $b_1 = 6$, then $d_1 \cdot 7 \cdot (49 - 18) = 1911$, but $49 - 18 = 31$ is not a divisor of $1911$.
If $a_1 = 2$, $b_1 = 5$, we have $d_1 \cdot 7 \cdot (49 - 30) = 1911$, but $49 - 30 = 19$ is not a divisor of $1911$.
Finally, if $a_1 = 3$, $b_1 = 4$, then $d_1 \cdot 7 \cdot (49 - 36) = 1911$, which gives $d_1 = 21$, whence $d = 21 \cdot 3 \cdot 4 = 252$, i.e., the pair $a = 756$, $b = 1008$ is a solution.

3. $a_1 + b_1 = 13$. If $a_1 = 1$, $b_1 = 12$, then $d_1 \cdot 13 \cdot (169 - 36) = 1911$, but $169 - 36 = 133$ does not divide $1911$.
If $a_1 = 2$, $b_1 = 11$, then $d_1 \cdot 13 \cdot (169 - 66) = 13 d_1 \cdot 103$ does not divide $1911$.
If $a_1 = 3$, $b_1 = 10$, then $d_1 \cdot 13 \cdot (169 - 90) = 13 d_1 \cdot 79$ does not divide $1911$.
If $a_1 = 5$, $b_1 = 8$, then $d_1 \cdot 13 \cdot (169 - 120) = d_1 \cdot 13 \cdot 49 = 1911$, which gives $d_1 = 3$, whence $d = 3 \cdot 5 \cdot 8 = 120$, i.e., the pair $a = 600$, $b = 960$ is a solution.
Finally, if $a_1 = 6$, $b_1 = 7$, then $d_1 \cdot 13 \cdot (169 - 126) = 13 d_1 \cdot 43$ does not divide $1911$.

Thus, the equation has two solutions mentioned above.