(I) To prove: Since $a_{n+2}=3a_{n+1}-2a_n$, we have
$a_{n+2}-a_{n+1}=2(a_{n+1}-a_n)$.
Because $a_1=1$ and $a_2=3$, we get
$$\frac{a_{n+2}-a_{n+1}}{a_{n+1}-a_n}=2 \text{ for } n\in \mathbb{N}^{*}.$$
Thus, $\{a_{n+1}-a_n\}$ is a geometric sequence with the first term $a_2-a_1=2$ and a common ratio of $2$.

(II) We know from (I) that $a_{n+1}-a_n=2^n$ for $n\in \mathbb{N}^{*}$, therefore,
$a_n=(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+\cdots+(a_2-a_1)+a_1$
$=2^{n-1}+2^{n-2}+\cdots+2+1$
$=2^n-1 \text{ for } n\in \mathbb{N}^{*}.$
So, the general term formula for the sequence is $\boxed{a_n=2^n-1}$.

(III) To prove: Since $4^{b_1-1} \cdot 4^{b_2-1} \cdots 4^{b_n-1} = (a_n+1)^{b_n}$, we have
$4^{b_1+b_2+\cdots+b_n-n}=2^{2(b_1+b_2+\cdots+b_n-n)}=2^{nb_n}$.
Thus, $2[(b_1+b_2+\cdots+b_n)-n]=nb_n$ (Equation 1).
For the next term in the sequence, we have
$2[(b_1+b_2+\cdots+b_n+b_{n+1})-(n+1)]=(n+1)b_{n+1}$ (Equation 2).
Subtract Equation 1 from Equation 2 to get
$2(b_{n+1}-1)=((n+1)b_{n+1}-nb_n)$,
which simplifies to $(n-1)b_{n+1}-nb_n+2=0$ (Equation 3).
For the $(n+2)$-th term, we have
$nb_{n+2}-(n+1)b_{n+1}+2=0$ (Equation 4).
Subtracting Equation 3 from Equation 4, we get
$nb_{n+2}-2nb_{n+1}+nb_n=0$,
which simplifies to $b_{n+2}-2b_{n+1}+b_n=0$.
Therefore, $b_{n+2}-b_{n+1}=b_{n+1}-b_n$ for $n\in \mathbb{N}^{*}$,
indicating that $\{b_n\}$ is an arithmetic sequence. Thus, the proof is complete.