Induction Proof $k^2 times 2^k$

I need help on this proof. I am not able to do it after setting m=m+1.

Prove by induction on n that sum of $k^2 \times 2^k$ from $k=1$ to $n$ is equal to $(n^2-2n+3) \times 2^{n+1}-6$

Base case:

Let $k=1$ so L.H.S side is $2$
Let $n=1$ so R.H.S side is $2$

Inductive hypothesis:

Let $n=m$ so $(m^2-2m+3) \times 2^{m+1}-6$

Proof:

Let $n=m+1$ so prove that $((m+1)^2-2(m+1)+3) \times 2^{m+2}-6=(m^2-2m+3) \times 2^{m+1}-6$

But I am not able to prove that they are equal.