First I showed that $\triangle u(x,y) = \partial_{xx} u+ \partial_{yy} u$ under the change of variable $z=x+iy, \bar z=x-iy$ we have $\triangle \tilde u(z,\bar {z}) = 4\partial_{\bar z}\partial_{{z}} \tilde u$.
Now given a real valued function $u(x,y)$ such that $\triangle^2 u = 0$, then I want to show for some real valued function $v(x,y)$, we have $$u(x,y)+iv(x,y) = \bar{z} f(z) + g(z)$$ where $f, g$ are analytic functions and $z=x+iy$.
My attempt: First, given $\triangle^2 u = 0$, we know $\triangle u$ is harmonic, so there exists a harmonic conjugate $\varphi$ such that $$\triangle u + i\varphi = h$$ where $h$ is some analytic function. So now we look for a function such that $\triangle F(z,\bar z) = h(z)$. From $$\triangle F =4\partial_{\bar z}\partial_{{z}} F(z,\bar z) = h(z)$$ we have $$4\partial_{{z}} F = \bar{z}h(z) + g(z)$$ $$4 F(z, \bar z) = \bar{z}H(z) + G(z) + l(\bar z).$$ So here, $H, G$ are both holomorphic function, but I have the extra term $l(\bar z)$ where $l(\bar z)$ is anti-holomorphic.
You are essentially done. Since the real part is taken, you can put a conjugate on $l(\bar z)$, making it holomorphic, and absorb it into $G$. Formally, $$ 4 F(z, \bar z) = \operatorname{Re}\left( \bar{z}H(z) + G(z) + l(\bar z)\right) = \operatorname{Re}\left( \bar{z}H(z) + G(z) + \overline{l(\bar z)}\right) = \operatorname{Re}\left( \bar{z}f(z) + g(z)\right) $$ where $f(z)=H(z)$ and $g(z) = G(z) + \overline{l(\bar z)}$.