Example 3 Find all integer solutions $(a, b)$ of the equation
$$
a^{4}-3 a^{2}+4 a-3=7 \times 3^{b}
$$
Given the equation factorizes to
$$
\left(a^{2}+a-3\right)\left(a^{2}-a+1\right)=7 \times 3^{b} \text {. }
$$

Clearly, $b \geqslant 0$ (otherwise it would contradict the left side being an integer).
Also, $\left(7,3^{b}\right)=1$, so we have
$\left\{\begin{array}{l}a^{2}-a+1=3^{x}, \\ a^{2}+a-3=7 \times 3^{y}, \\ x+y=b,\end{array}\right.$
or $\left\{\begin{array}{l}a^{2}-a+1=7 \times 3^{m}, \\ a^{2}+a-3=3^{n}, \\ m+n=b .\end{array}\right.$
First, consider the system of equations (1).
For $a^{2}-a+1=3^{x}$ to have integer solutions,
$$
\Delta_{1}=1-4+4 \times 3^{x}=4 \times 3^{x}-3
$$

must be a perfect square (otherwise the quadratic formula would have a square root).
When $x \geqslant 2$, $3 \mid \Delta_{1}$, but $9 \nmid \Delta_{1}$, so $\Delta_{1}$ is not a perfect square.
Checking $x=0,1$, there are no solutions.
Next, consider the system of equations (2).
For $a^{2}-a+1=7 \times 3^{m}$ to have integer solutions,
$$
\Delta_{2}=28 \times 3^{m}-3
$$

must be a perfect square.
When $m \geqslant 2$, $3 \mid \Delta_{2}$, but $9 \nmid \Delta_{2}$, so $\Delta_{2}$ is not a perfect square.
When $m=0$, $a=3, b=2$.
When $m=1$, $a=5, b=4$ or $a=-4, b=3$.
In summary, the solutions satisfying the conditions are $(a, b)=(3,2),(5, 4), (-4,3)$.