What you really want to show is that the function $f(n)=2^n-1$ that expresses your closed form solution satisfies the recurrence and initial conditions of the Tower of Hanoi problem. The recurrence should be $T_n=2T_{n-1}+1$ for $n>0$. You want to prove by induction that $T_n$, defined by this recurrence with initial condition $T_1=1$, is equal to $f(n)$ for every positive integer $n$.
First we get the induction off the ground: $f(1)=2^1-1=1=T_1$, so all’s well. Now assume that $T_n=f(n)$ for some $n\ge 1$; that’s your induction hypothesis, and you want to use it to prove that $T_{n+1}=f(n+1)$. The computation is pretty straightforward:
$\begin{align*} T_{n+1}&\overset{(1)}=2T_n+1\\ &\overset{(2)}=2f(n)+1\\ &\overset{(3)}=2\left(2^n-1\right)+1\\ &\overset{(4)}=2\cdot2^n-2+1\\ &\overset{(5)}=2^{n+1}-1\\ &\overset{(6)}=f(n+1) \end{align*}$
Here $(1)$ is by virtue of the recurrence definining the numbers $T_k$, $(2)$ uses the induction hypothesis that $T_n=f(n)$, $(3)$ replaces $f(n)$ by its definition, $(4)$ and $(5)$ are algebra, and $(6)$ is the definition of $f$ again.
Since you’ve checked that $T_1=f(1)$ and that $T_n=f(n)$ implies that $T_{n+1}=f(n+1)$, it follows from the principle of mathematical induction that $T_n=f(n)=2^n-1$ for all positive integers $n$.
Of course this is an excessively careful presentation of the argument; in practice both the calculation and the verbiage can be reduced a bit.