2. Note that the coefficients of the equation 3 and 453 are divisible by 3, then the term $5 y^{2}$ is also divisible by 3, so $y=3 t$, $y^{2}=9 t^{2}$, where $t \in \mathbb{Z}$. The equation takes the form: $3 x^{2}+5 \cdot 9 t^{2}=453, x^{2}+5 \cdot 3 t^{2}=151, x^{2}=151-15 t^{2}, \quad$ where $\quad x \in \mathbb{Z}$. Substitute possible values of $t \geq 0$.

| Values of $t$ | $15 t^{2}$ | $x^{2}=151-15 t^{2}$ | Integer value of $x$. |
| :--- | :--- | :--- | :--- |
| $t=0$ | 0 | $x^{2}=151$ | no |
| $t=1$ | 15 | $x^{2}=151-15, x^{2}=136$ | no |
| $t=2$ | 60 | $x^{2}=151-60, x^{2}=91$ | no |
| $t=3$ | 135 | $x^{2}=151-135, x^{2}=16$ | $x= \pm 4$ |
| $t=4$ | 240 | $x^{2}=151-240<0$ | no |

Thus, $x= \pm 4, t= \pm 3, y= \pm 9$.
Answer: 4 pairs, $( \pm 4 ; \pm 9)$.