Let $V$ be a vector space. Let $T$ and $S$ be in $L(V)$ and satisfy $ST-TS=I$ where $I$ is identity operator. We want to prove that $L(V)$ is infinite-dimensional. My try is:
Using the given condition we can show using induction $S^n T - TS^n=(n+1)S^n$ where $n$ is non negative integer. Now a vector space is infinite-dimensional if and only if it has a infinite linearly independent subset. Here I am stuck. Thanks.
Suppose that $V$ is a vector space of dimension $n$ over a field $F$ of characteristic not dividing $n$. If there are linear operators $S,T$ on $V$ such that $ST-TS=I$, then taking the trace of both sides leads to $$ 0=\mathrm{tr}(ST-TS)=\mathrm{tr}(I)=n $$ which is a contradiction. Hence no such $S$ and $T$ can exist.
In particular, if $F$ has characteristic zero then we can conclude that if there are linear operators $S$ and $T$ on $V$ such that $ST-TS=I$, then $V$ must be infinite dimensional.
However, if $F$ has non-zero characteristic then it is possible to have $ST-TS=I$ when $V$ is finite-dimensional. Consider for instance $F=\mathbb{F}_2$, $V=F^2$, and $$ S=\begin{bmatrix}0&1\\0&0\end{bmatrix}$$ and $$ T=\begin{bmatrix}0&0\\1&0\end{bmatrix} $$ Then $$ ST-TS=ST+TS=\begin{bmatrix}1&0\\0&0\end{bmatrix}+\begin{bmatrix}0&0\\0&1\end{bmatrix}=I $$ since $1=-1$ in $F$.
This answer may include some ideas you're not familiar with, but it does provide a proof. Also, I will assume the vector space is over $\mathbb R$ or $\mathbb C$.
Suppose $V$ is finite-dimensional (which occurs if and only if $L(V)$ is) then it can be given the structure of a normed vector space (by choosing a basis then choosing an $\ell^p$-norm). Since $V$ is normable, $L(V)$ can be instilled with the corresponding operator norm, making it a unital normed algebra. From here, there is a well-known result that in any unital normed algebra $A$ with unit $I$, then $ST-TS=I$ cannot be satisfied by any $S,T\in A$ (i.e. the identity is not a commutator).
By contraposition, we obtain the result.