The corrections arising from the Schwarzschild solution, compared to the dynamics given by Newtonian gravity, are small, but apply to all bodies orbiting the sun, not just Mercury. But, Mercury is the flightiest of the planets - which is why it got its name - and the correction happens to be proportional to the square of the "areal velocity" (i.e. angular momentum per unit mass) and inversely proportional to the square of the distance from the sun. It's close to the sun and it's whipping around pretty fast, so it's got the biggest correction.
Without going through the derivation in detail (instead, I derive it here Modified Kepler Equation), the equations for orbital dynamics: $$\frac{d^2𝐫}{ds^2} = -\frac{μ𝐫}{r^3},\quad \frac{dt}{ds} = 1,\label{0}\tag{Newtonian}$$ where $μ = G M$, with $G$ being Newton's coefficient, $𝐫$ the position vector centered on the Schwarzschild object (here: the sun), $r = |𝐫|$ is the distance to the object, and $M$ the mass of the object, are corrected to: $$\frac{d^2𝐫}{ds^2} = -\frac{μ𝐫}{r^3}\left(1 + \frac{3α}{r^2}\left|𝐫×\frac{d𝐫}{ds}\right|^2\right),\quad \frac{dt}{ds} = \frac{Ar}{r - 2αμ}\label{1}\tag{Einsteinian},$$ where the "time-warping" factor $A$ is determined as the positive solution to the equation $$1 = \frac{r - 2αμ}{r} \left(\frac{dt}{ds}\right)^2 - α \left(\left|\frac{d𝐫}{ds}\right|^2 + \frac{2αμ}{r - 2αμ} \left(\frac{dr}{ds}\right)^2\right).$$ This equation and the (\ref{1}) equation hold for "time-like" geodesics; i.e. orbital trajectories for sub-light-speed bodies.
The correction is parametrized by $α = 1/c^2$, where $c$ is light speed. In the limit $c → ∞$, the parameter $α → 0$, and the equation (\ref{1}) → (\ref{0}).
The areal velocity is given by the vector $$𝐡 = 𝐫×\frac{d𝐫}{ds},$$ which is a constant of motion for the equations above and is just $h = |𝐡|$. Using it, you can write $$\left|\frac{d𝐫}{ds}\right|^2 = \left(\frac{dr}{ds}\right)^2 + \frac{h^2}{r^2},$$ and $$\frac{d^2𝐫}{ds^2} = -\frac{μ𝐫}{r^3}\left(1 + 3α\frac{h^2}{r^2}\right).$$ The equation for the time-warping factor $A$ reduces to: $$1 = \frac{A^2 r}{r - 2αμ} - α \left(\frac{r}{r - 2αμ}\left(\frac{dr}{ds}\right)^2 + \frac{h^2}{r^2}\right),$$ or, taking the positive root: $$A = \sqrt{\frac{r - 2αμ}{r}\left(1 + α\frac{h^2}{r^2}\right) + α \left(\frac{dr}{ds}\right)^2}.$$