Are $\mathsf{\#P}$ problems harder than $\mathsf{NP}$ problems

I have a method to solve the $ \mathsf{\#P}$ version of 3SAT in a way that seemingly reduces it to an $ \mathsf{NP}$ problem. – I don’t have a formal understanding of these terms so I will just show an example here.

How many solutions does $ \phi=\left(x_1 \vee x_2 \right) \wedge \left(\neg x_1 \vee x_3 \right) \wedge \left(\neg x_2 \vee \neg x_3 \right)$ have?

First convert to a boolean polynomial: $ $ \phi=\left(x_1 + x_2 -x_1x_2\right) \left(1-x_1+x_1x_3\right) \left(1-x_2x_3 \right)$ $

Expand and simplify: $ $ \phi = x_2 – x_1x_2 + x_1x_3 – x_2x_3$ $

Substitute $ x_1 = x_2 = x_3 = \frac{1}{2}$ and multiply by $ 2^n$ where $ n$ is the number of variables: $ $ |\phi| = 2^n\left(\frac{1}{2} – \frac{1}{4} + \frac{1}{4} – \frac{1}{4}\right) = 2^3\frac{1}{4} = 2$ $

where $ |\phi|$ is the number of solutions that $ \phi$ has.

And indeed, those 2 solutions are (0,1,0) and (1,0,1).

I have proven that this works for any logical expression. Does this have any use/bearing on $ \mathsf{\#P}$ vs $ \mathsf{NP}$

Thanks in advance, Ben.