Given an algebra $\mathfrak{g}$, the killing form is defined as $K(x, y) = \operatorname{Tr}(\operatorname{ad}(x) \circ \operatorname{ad}(y))$, but when $\mathfrak{g}=\mathfrak{gl}(n)$, we have that:
$\operatorname{Tr}([x, y])=\operatorname{Tr}(xy-yx)=0$
so is it true that $\operatorname{Tr}([\mathfrak{gl}(n), \mathfrak{gl}(n)])=0?$
The Killing form of $\mathfrak{gl}_n(K)$ can be expressed by the trace form, i.e., we have $$ B(X,Y)=2n \cdot \operatorname{tr}(XY) − 2 \operatorname{tr}(X)\operatorname{tr}(Y) $$ For $[\mathfrak{gl}_n(K), \mathfrak{gl}_n(K)]=\mathfrak{sl}_n(K)$ the Killing form therefore is given by $$ B(X,Y)=2n \cdot \operatorname{tr}(XY) $$ since $\operatorname{tr}([A,B])=\operatorname{tr}(AB-BA)=0$, as you said. Although this trace is zero, the Killing form of $\mathfrak{sl}_n(K)$ is not necessarily zero (but it can be for prime characteristic $p$ with $p\mid n$). I hope this helps with the misunderstanding.
Reference:
Uniqueness of the killing form