The law of gravitation is
$$F=-\frac{GMm}{r^2}$$
Where $F$ is the force exerted by $M$ on $m$ and vice versa. The gravitational potential energy function can be derived from the law of gravitation.
Lets assume $M$ is the mass of the earth and $m$ is your mass. The force the earth exerts on you will equal your mass, $m$, times the acceleration due to gravity at the surface of the earth, $g$. We can calculate $g$ from the following
$$g=-\frac{GM}{r^2}$$
Where $G$ is the universal gravitational constant = $6.674x10^{11} N.kg^{-2}.m^2$
$M$ is the mass of the earth = $5.9742 x10^{24}kg$
$r$ is the radius of the earth = $6378 km$ at the equator.
- $G$ is the universal gravitational constant $= 6.674\times10^{11}$ N kg$^{-2}$ m$^2$
- $M$ is the mass of the earth $= 5.9742 \times 10^{24}$ kg
- $r$ is the radius of the earth $= 6378$ km at the equator.
Which gives us, using proper units, $g=-9.8 \frac{m}{s^2}$$g=-9.8 \frac{\text{m}}{\text{s}^2}$ at the surface of the earth.
Now, when we apply the equation $GPE=mgh$, it is the difference in the gravitational potential energy of $m$ between the surface of the earth and the height $h$. $g$ can be assumed to be constant provided $h$ is negligibly small compared to $r$. If it isn’t, then you would need to redo the above calculation to determine a new $g$ using a new value of $r$
Bottom line, the $h$ in $mgh$ is not the same as $r$ in the gravitational potential energy equation. Consequently an increase in either $m$ or $h$ results in an increase in gravitational potential energy near the surface of the earth. This does not contradict the gravitational potential energy function.
Hope this helps.
