For simplicity consider $f \in \mathcal{S}$. Fix $\psi\in C^\infty_c(\mathbb{R}^n,[0,\infty))$ with $\int \psi = 1$. Define
$$ \varphi_{\epsilon,y}(x) = \epsilon^{-n}\psi( \epsilon^{-1}(x-y)) $$
so we have $\int\varphi_{\epsilon,y} = 1$ for every $y\in \mathbb{R}^n$ and $\epsilon > 0$.
If the desired inequality were to hold, you must have
$$ |\varphi_{\epsilon,y}* f(x)| \lesssim |Mf(x)| $$
for your choice of maximal function $M$. Taking $\epsilon \searrow 0$ the LHS converges (as we are using an approximation to identity), and so we must have
$$ |f(x+y)| \lesssim |Mf(x)| $$
uniformly for $y\in \mathbb{R}^n$. Taking the supremum in $y$ we find necessarily
$$ \|f\|_\infty \lesssim |Mf(x)|. $$
In particular, we have that for any non-trivial function, $|Mf|$ is bounded below away from zero. As consequence, we can conclude that
- If $M$ is any maximal function that maps $L^p$ to itself boundedly, for some $1 \leq p < \infty$, then the desired inequality cannot hold.
- In fact, if there is some $p\in [1,\infty]$ and $q\in [1,\infty)$ such that $M$ sends all $L^p$ functions to $L^{q}_w$ (weak $L^q$) functions, then the desired inequality cannot hold.
- In a slightly different direction: if $M$ is any maximal function that maps $L^p$ to $L^\infty$ for some $1\leq p < \infty$, then the desired inequality cannot hold.
The astute reader will notice that the properties that appear in the hypotheses of the above three statements are the three quintessential properties of the Hardy-Littlewood Maximal function, and some subset thereof are often shared by what one consider to be "maximal functions".
