Example 1 Try to express the polynomial $x^{4}+3 x^{2}-2 x+3$ as the difference of squares of two polynomials with integer coefficients of different degrees, and find all solutions.

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Analysis: If $x^{4}+3 x^{2}-2 x+3=f^{2}-g^{2}, f$, $g$ are polynomials with integer coefficients, $\operatorname{deg} f \neq \operatorname{deg} g$, then $(-f, g)$, $(f,-g),(-f,-g)$ are also solutions. Therefore, we can assume that the leading coefficients of $f$ and $g$ are both positive. Thus, we can set $f(x)=x^{2}+a x+b, g(x)=c x+d$ to satisfy the conditions, where $a, b, c, d$ are undetermined coefficients, and $c>0, a, b, c, d$ are integers. Therefore, $f^{2}-g^{2}=\left(x^{2}+a x+b\right)^{2}-(c x+d)^{2}$, so
$$
x^{4}+3 x^{2}-2 x+3=x^{4}+2 a x^{3}+\left(a^{2}+2 b-\right.
$$
$\left.c^{2}\right) x^{2}+(2 a b-2 c d) x+b^{2}-d^{2}$.
By the polynomial identity theorem, we get the system of equations
$$
\left\{\begin{array}{l}
2 a=0, \\
a^{2}+2 b-c^{2}=3, \\
2 a b-2 c d=-2, \\
b^{2}-d^{2}=3 .
\end{array}\right.
$$

Solving, we get $a=0, b=2, c=1, d=1$. Therefore, $f(x)=$ $x^{2}+2, g(x)=x+1$.

The other three solutions are $-f(x), g(x) ; f(x)$, $-g(x) ;-f(x),-g(x)$.