Show that $\widehat{Cov(\hat{\mu},Z_i)}$ is always zero even $Cov(\mu,Z_i)$ is not always $0$

I will state the question first then my work.

Q:

We have a regression model, $Y_i=\beta_0+\beta_1X_i+\mu_i$ where $Cov(\mu_i,X_i)=0$ is not guaranteed. Suppose that $Z_i$ is an instrumental variable and we have $\hat{\beta_1}$ be the two least squares estimator and we have $\hat{\beta_0}=\bar{Y_n}-\beta_1\bar{X_n}$. Show that $\widehat{Cov(\hat{\mu},Z_i)}$ is always zero even $Cov(\mu,Z_i)$ is not always $0$.

My thoughts and work:

We know that the residual is $\hat{\mu_i}=Y_i-\hat{\beta_0}-\hat{\beta_1}X_i$ so...

$$\widehat{Cov(\hat{\mu},Z_i)}=\widehat{Cov(Y_i,Z_i)}-\widehat{Cov(\beta_0,Z_i)}-\widehat{Cov(B_1X_i,Z_i)}$$

and thus the second term would be equal to $0$ since $B_0$ doesn't have an $i$ term. But then I'm not sure how to proceed IF this is the right way to move forward.

Step by step help would be greatly appreciated!

When you use instrumental variables via two stage least squares, for instance, you have a first stage and a second stage respectively: \begin{align} X_i &= \delta_0 + \delta_1 Z_i + \eta_i & \text{first stage}\newline Y_i &= \beta_0 + \beta_1 X_i + \mu_i & \text{second stage} \end{align}
A possible solution path would be to plug the first stage into the second stage and then use the fact that an OLS regression of this equation separates the variation in $Y_i$ into an observed part and an unobserved part. Using linear projections you can then show why this observed part (your $Z$ in this case) will be orthogonal to $\mu_i$. This is basically a purely mechanical effect from the regression (see here for an explanation), hence the estimated covariance between $Z_i$ and $\mu_i$ is going to be zero.
The next step would be to make an argument for why the population covariance $Cov(Z_i,\mu_i)\neq 0$. This basically aims at the point as to why two stage least squares could be biased. Note though that there is no test for whether $Cov(Z_i,\mu_i)=0$ in the population or not - it is always made as an assumption but it can never be tested.