Let $\mathcal{B}(\mathcal{H})$ the algebra of all bounded linear operators on an infinite dimensional complex Hilbert space $\mathcal{H}$.
Let $T,S\in \mathcal{B}(\mathcal{H})$. I want to prove that the equality \begin{equation}\label{commz} [T,S]:=TS-ST=I \tag{1}, \end{equation} cannot hold.
To see this, assume that $(1)$ holds. We shall prove by induction that \begin{equation}\label{tag2} [T, S^n] = nS^{n - 1},\;n\in \mathbb{N}^*. \end{equation} By assumption, $[T,S]=S^{0}=I$. Suppose $[T,S^{n}]= nS^{n-1}$ for some $n\in \mathbb{N}^*$. Then \begin{align*} [T,S^{n+1}] & = TS^{n+1}-S^{n+1}T\\ & =(TS^{n}-S^{n}T)S+S^{n}TS-S^{n+1}T \\ & = [T,S^{n}]S+S^{n}[T,S] \\ & = nS^{n-1}S+S^{n}=(n+1) S^{n}. \end{align*} So, \begin{equation*}\label{tag11} TS^n - S^nT = n S^{n=1}, \end{equation*} holds for all $n\in \mathbb{N}^*$. Hence, \begin{align*} n\|S^{n-1}\| & = \|TS^n - S^nT\|\\ &\leq 2 \|T\|\cdot\|S^{n} \|\\ &\leq 2 \|T\|\cdot\|S\|\cdot\|S^{n-1} \|. \end{align*}
If $\|S^{n-1} \|\ne 0$, we get $n \le 2 \|T\|\|S\|$, for all $n\in \mathbb{N}^*$. This leads to a contradiction. So, $(1)$ cannot hold for $T,S\in \mathcal{B}(\mathcal{H})$.
Why $S^{n-1}\ne 0$ for all $n\in \mathbb{N}^*$?
Indeed, the conclusion holds only if $S^{n-1}\ne 0$.
But, assume $S^k=0$ for some $k$, then by your formula $[T, S^k] =kS^{k-1}$, we get $S^{k-1}=0$, which in turn implies $S^{k-2}=0$, and so on, until $S=0$ which is impossible, provided that $[T, S] =I$.