I found the right reference, and read them and carry them properly to graded case. 

The main lemma is [Auslander-Buchsbaum][1]'s argument

> Let $R$ be a **noetherian ring**, 
$M$ a **finitely generated** module. Assume $M$ admit a **finite finitely generated free resolution**.
Then if the annihilator of $M$ is not trivial, then it contains a nonzero divisor in $R$.

The sketch of the proof is as the following. 

 1. Firstly, show that $M_{\mathfrak{p}}$ is free for any associaed prime $\mathfrak{p}$ of $R$. This is essentially the main process of [Auslander-Buchsbaum equality][2], but one can use matrix coefficient trick to prove it directly. 
--- Here we use the assumption that $R$ is noetherian. 

 2. But the annihilator $\mathfrak{a}$ kills $M_{\mathfrak{p}}$, so $\mathfrak{a}_{\mathfrak{p}}=0$ or $M_{\mathfrak{p}}=0$. 
Since the rank to get the rank of $M_{\mathfrak{p}}$ does not depend on $\mathfrak{p}$. 
--- Here we use the assumption of finiteness of free resolution. 

 3. But if $\mathfrak{a}_{\mathfrak{p}}=0$, then the annihilator of $\mathfrak{a}$ is not contained in any prime associated to $R$, so their union, the zero divisor. Thus $\mathfrak{a}=0$. 
--- Here we use the assumption that $R$ is noetherian. 

 4. So $M_{\mathfrak{p}}=0$, then then the annihilator of $\mathfrak{a}$ is not contained in any prime associated to $R$, so their union, the zero divisor. 
--- Here we use the assumption that $M$ is finitely generated. 

So we are done. 

> For a polynomial ring $R$ over field,
the homogenous ideal $I$ is generated by a regular sequence of homogenous elements if $I/I^2$ is free over $R/I$.

The sketch of the proof is. the following.  

 1. Note that $I$ admits a finite finitely generated free (twisted) resolution due to [Quillen–Suslin theorem][3]. 
--- Since we only need an existence, maybe the one who do not want to use such big theorem can use only Scheja-Stroch's computational proof of Hilbert’s syzygy theorem (for example, Weibel page 114).  

 2. As the annihilator of $R/I$, it consists some non zero divisor. So is $I\setminus R_+ I$ by [(strong) prime avoidance][4] and the fact $I\neq R_+I$. 

 3. Pick such nonzero divisor $x\in I\setminus R_+ I$, then consider $\overline{R}=R/xR$, and $\overline{I}$ the image of $I$. 

 4. It is clear, now $\overline{I}/\overline{I}^2=I/(I^2+xR)$ is free of less rank than $\overline{R}/\overline{I}=R/I$. 
--- Here we use that it is over some field, and graded, otherwise, one cannot claim like this, since $x$ may not extend to a basis of $I/I^2$ over $R/I$. 

 5. $\overline{I}$ admits a finite finitely generated free (twisted) resolution. 
--- remind the prove $pd_R M=pd_{R/xR} M/xM$ for non zero divisor $x$ for both $R$ and $M$. 

Then it follows from induction. 

The above process follows for neotherian local ring, as done in [Ideals generated by R-sequences][5]. 

But it is also not clear what will happen for general graded ring. 

----------------------------------------------------

Edit: My classmate remind me that this is also true. 

> Let $R$ be a **connected noetherian ring**, 
$M$ a **finitely generated** module. Then if the annihilator of $M$ is not trivial, then it contains a nonzero divisor in $R$.

By the same way. So we also have this 

> For a graded ring $R$ over field $k$ with $R^0=k$, only nonnegative degree, 
a homogenous ideal $I$ is generated by a regular sequence of homogenous elements if $I/I^2$ is free over $R/I$ and $I$ has finite projective dimension.

Besides, for any set of element presenting a basis, by our choice, our choice of regular sequence presents a set of basis of $I/I^2$ over $R/I$, so it differs by our choice a invertible matrix. Then it reduces to exchange two element. We permute them by degree reason. So in conclusion

> In above case, any set of basis presenting a set of basis for $I/I^2$ over $R/I$ forms a regular sequence. 

  [1]: https://stacks.math.columbia.edu/tag/090V
  [2]: https://stacks.math.columbia.edu/tag/090U
  [3]: https://en.wikipedia.org/wiki/Quillen%E2%80%93Suslin_theorem
  [4]: https://stacks.math.columbia.edu/tag/00DS
  [5]: https://core.ac.uk/download/pdf/82227314.pdf