On the disk $\{z:|z|<R\}$, Poisson's integral formula is $$u(r,\theta)=\frac1{2\pi}\int_0^{2\pi}\frac{(R^2-r^2)f(\phi)}{R^2-2Rr\cos(\theta-\phi)+r^2}\,d\phi$$ which solves the Dirichlet problem. Taking $R=1$ for simplicity, the Fourier series of $u$ is obtained through $$\begin{align}u(r,\theta)&=\frac1{2\pi}\int_0^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\phi)+r^2}f(\phi)\,d\phi\tag1\\&=\frac12a_0+\sum_{n=1}^\infty(a_n\cos n\theta)r^n+\sum_{n=1}^\infty(b_n\sin n\theta)r^n\tag2\end{align}$$ where $\pi a_n=\int_0^{2\pi}f(\phi)\cos n\phi\,d\phi$ and likewise for $b_n$. Now consider the generating function of the Chebyshev polynomials $T_n$ given by $$\frac{1-rx}{1-2rx+r^2}=\sum_{n=0}^\infty T_n(x)r^n\implies\frac{1-r\cos\theta}{1-2r\cos\theta+r^2}=1+\sum_{n=1}^\infty(\cos n\theta)r^n\tag3$$ on substituting $x:=\cos\theta$. The LHS looks similar to $(1)$ and the RHS to $(2)$ but for the factor of $a_n$.
Is there a direct link between the the generating function of $T_n$ and Poisson's integral formula; that is, can $(3)$ be derived from $(1)$?
Not sure if that is what you want but by definition the Poisson kernel is:
$P(z)=\sum_{n <0}r^{|n|}e^{in\theta}+\sum_{n \ge 0}r^{n}e^{in\theta}=1+2\sum_{n \ge 1}r^n \cos \theta, z=re^{i\theta}$
while
$T(z)=\sum_{n \ge 0} T_n(x)r^n=\sum r^n \cos \theta$ since $T_n(x)=\cos n\theta, x=\cos \theta, z=re^{i\theta}$,
so $2T(z)=P(z)+1$ and obviously the relations will look similar beacuse $P,T$ are related in this simple way