Example 1 For all real numbers $p$ satisfying $0 \leqslant p \leqslant 4$, the inequality $x^{2}+p x>4 x+p-3$ always holds. Try to find the range of values for $x$.
Analysis We take $p$ as the main variable, transforming it into a linear function problem, thereby simplifying the calculation.
Solution The original inequality is equivalent to $p(x-1)+\left(x^{2}-4 x+3\right)>0$. Let $f(p)=(x-1) p+\left(x^{2}-4 x+3\right)$, then $f(p)$ is a linear function of $p$. It is easy to see that the original inequality is equivalent to $\left\{\begin{array}{l}f(0)>0 \\ f(4)>0\end{array}\right.$, which is $\left\{\begin{array}{l}x^{2}-4 x+3>0 \\ x^{2}-1>0\end{array}\right.$. Solving this, we get $x>3$ or $x<-1$.