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, without loss of generality, we can assume that the leading coefficients of $f, g$ are both positive. Thus, we can set $f(x)=x^{2}+a x+b, g(x)=c x+d$.

Solution Without loss of generality, let $f(x)=x^{2}+a x+b, g(x)=c x+d$ satisfy the conditions, where $a, b, c, d$ are undetermined coefficients, and $c>0, a, b, c, d$ are integers. Thus, $f^{2}-g^{2}=\left(x^{2}+a x+b\right)^{2}-(c x+d)^{2}$, so $x^{4}+3 x^{2}-2 x+3=x^{4}+2 a x^{3}+\left(a^{2}+2 b-c^{2}\right) x^{2}+(2 a b-2 c d) x+b^{2}-d^{2}$. By the identity of polynomials, 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 $(-f, g),(f,-g),(-f,-g)$.