There is a relationship between these two concepts. Let the complex function $f(z)$ be analytic on and inside a simple closed curve $C$ in the complex plane. Then Cauchy's integral formula states that for any point $z$ inside $C$ we have
$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=2\pi j f(z)\tag{1}$$
For $z$ outside $C$ we have
$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=0\tag{2}$$
Furthermore, for $z$ on the curve $C$ the following holds:
$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=\pi j f(z)\tag{3}$$
A proof of Eq. $(3)$ can be found in any text on complex analysis.
Let's now choose the curve $C$ as the real line from $-R$ to $R$ and the semi-circle $Re^{j\phi}$, $\phi\in[0,\pi]$, such that $C$ is closed. For $R\to\infty$, the curve $C$ "encloses" the upper half-plane. If $f(z)$ is analytic in the upper half-plane, and if it decays to zero sufficiently fast such that the contribution of the integral along the semi-circle with radius $R$ vanishes as $R\to\infty$, Eq. $(3)$ becomes
$$\int_{-\infty}^{\infty}\frac{f(\zeta)}{\zeta -x}d\zeta=\pi j f(x)\tag{4}$$
Note that the integral in $(4)$ is along the real line. With $z=x+jy$ and $f(z)=f_R(x,y)+jf_I(x,y)$, Eq. $(4)$ can be written as
$$\int_{-\infty}^{\infty}\frac{f_R(\zeta,0)+jf_I(\zeta,0)}{\zeta -x}d\zeta=\pi j \big[f_R(x,0)+jf_I(x,0)\big]\tag{5}$$
Splitting Eq. $(5)$ into real and imaginary parts yields
\begin{align} f_I(x,0) & = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f_R(\zeta,0)}{\zeta-x}d\zeta &&= \mathscr{H}\big\{f_R(x,0)\big\} \tag{6}\\ f_R(x,0) & = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f_I(\zeta,0)}{\zeta-x}d\zeta &&= -\mathscr{H}\big\{f_I(x,0)\big\}\tag{7} \end{align}
where $\mathscr{H}\{\cdot\}$ denotes the Hilbert transform.
Equations $(6)$ and $(7)$ are identical to Eqs $(1)$ and $(2)$ in the question with $x_R(t)=f_R(t,0)$ and $x_I(t)=f_I(t,0)$. Consequently, the analytic function $f(z)$ equals an analytic signal on the real line.
Example 1:
Given the analytic signal $x(t)=e^{j\omega_0t}$, $\omega_0>0$, the corresponding analytic function is given by $f(z)=e^{j\omega_0z}$. Note that $f(z)$ satisfies all properties required for Eqs $(4)-(7)$ to be valid: it is analytic everywhere, and it decays rapidly for $|z|\to\infty$ and $\textrm{Im}\{z\}>0$ (i.e., in the upper half-plane).
Example 2:
We can also use functions with the required properties (analytic and decaying sufficiently fast in the upper half plane) to generate Hilbert transform pairs, or, equivalently, analytic signals. Let
$$f(z)=\frac{e^{j\omega_0 z}-1}{j\pi z},\qquad\omega_0>0\tag{8}$$
The function $f(z)$ is analytic everywhere (if we define $f(0)=\omega_0/\pi$), and it decays rapidly in the upper half plane $\textrm{Im}\{z\}>0$. Consequently, by evaluating $f(z)$ on the real line, we obtain an analytic signal and a Hilbert transform pair. For notational convenience, let's use $t$ to denote the real part of $z$. The corresponding analytic signal is given by
\begin{align} x(t) &= \frac{e^{j\omega_0 t}-1}{j\pi t} \\ & = \frac{\cos(\omega_0t)+j\sin(\omega_0t) - 1}{j\pi t} \\ & = \frac{\sin(\omega_0t)}{\pi t} + j\frac{1 - \cos(\omega_0t)}{\pi t} \tag{9} \end{align}
It has been shown in the answers to this question that the real and imaginary parts of $(9)$ indeed form a Hilbert transform pair. From the same answers, the Fourier transform of $x(t)$ can be easily derived:
$$X(j\omega)=\begin{cases}2,&0<\omega<\omega_0\\0,&\textrm{otherwise}\end{cases}\tag{10}$$
showing (again) that $x(t)$ is an analytic signal.
[To be continued .......]