Why do we need at least as many instrumental variables as endogenous regressors to identify parameters in 2SLS?

As the title says. Why do we need at least as many instrumental variables as endogenous regressors to identify parameters?

$$y = x^\top \beta + \epsilon$$
with exogeneity $$\mathbb E[x\epsilon] = \mathbf 0$$ you have $$K$$ parameters because $$\beta$$ is $$K \times 1$$ and you have $$K$$ equations
$$\mathbb E[x\epsilon] = \mathbb E[x(y - x^\top \beta)] = \mathbf 0 \Leftrightarrow \mathbb E[xy] = \mathbb E[xx^\top]\beta,$$
where $$\mathbb E[xy] = \mathbb E[xx^\top]\beta$$ is a system of linear equations. Only if the matrix $$\mathbb E[xx^\top]$$ is invertible it follows that $$\beta := \mathbb E[xx^\top]^{-1}\mathbb E[xy]$$.
For each endogenous variable in $$x_k$$ you lose one identifying equation because for the endogenous variable $$x_k$$ it does not hold that $$\mathbb E[x_k\epsilon] = 0$$. So for example with 2 endogenous variables you would be trying to identify $$K$$ parameters using only $$K-2$$ equations which amounts to trying to solve a system of $$K-2$$ linear equations for $$K$$ unknowns. So assuming that you know how systems of linear eqautions work it should come as no surprise that you need to find at least two instruments. This would allow using the true equation $$\mathbb E[z_k (y - x^\top \beta)] =0$$ instead of the untrue $$\mathbb E[x_k (y - x^\top \beta)] =0$$ for each $$k$$ (which amounts to assuming $$z_k$$ is exogenous while $$x_k$$ is not).