The problem states: If $a_{n+1}=3a_n-2a_{n-1}$ and $a_0=2$, $a_1=3$, prove by induction that $a_n=2^n+1$.
My strategy was to assume that $a_n=2^n+1$ indeed is true, then find expressions for both $a_{n+1}$ and $a_{n-1}$ and then show that from the assumption $a_{n+1}=3a_n-2a_{n-1}$ does follow.
Having checked the base case, let us assume that $a_n=2^n+1$, hence $a_{n+1}=2^{n+1}+1$ and $a_{n-1}=2^{n-1}+1$. Now let's write the needed equation and see whether we stumble upon some contradiction. $$a_{n+1}=3a_n-2a_{n-1}\Rightarrow 2^{n+1}+1=3\cdot2^n+3-2\cdot2^{n-1}-2$$ which leads to $$2^{n+1}+1=2^{n+1}+1$$ which indeed is true. Would this represent a correct inductive proof?
The only "mistake" is that you cannot deduce that the statement holds for $n+1$ from the fact that it holds for $a_n$. This is what you want to show. So, your mistake is exactly in the first sentence "assume that $a_n=2^n+1$, hence..." This "hence" is justified for $n-1$ but is not justified for $n+1$.
Formally, call the statement that you want to prove $P(k)$ (i.e. $P(k)$ states that $a_k=2^k+1$), and then
Hence, by the hypothesis $a_{n}=2^n+1$ and $a_{n-1}=2^{n-1}+1$, which gives $$a_{n+1}=3a_n-2a_{n-1}=3(2^n+1)-2(2^{n-1}+1)=2^{n+1}+1$$