Example 3 Given the equation in terms of $x$: $3^{2 x+1}+(m-1)\left(3^{x+1}-1\right)-(m-3) 3^{x}=0$ has two distinct real roots, find the range of real values for $m$.

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Let $3^{x}=t(t>0)$, the original equation can be transformed into: $3 t^{2}+(m-1)(3 t-1)-(m-3) t=0$, that is,
$$
3 t^{2}+2 m t-m+1=0
$$

The original problem is equivalent to the equation (1) having two distinct positive roots for $t$, then
$$
\left\{\begin{array}{l}
\Delta=4 m^{2}-4 \times 3(-m+1)>0 \\
t_{1}+t_{2}=-\frac{2}{3} m>0, \\
t_{1} \cdot t_{2}=\frac{1-m}{3}>0,
\end{array}\right.
$$

This yields $m<\frac{-3-\sqrt{21}}{2}$.
Key Points for Solving The key to solving unconventional exponential equations is transformation, converting them into simpler, more manageable exponential equations or applying function and equation ideas comprehensively. Transform the equation problem into a function problem and solve it using function ideas and methods, such as monotonicity and graphical methods.