The question was asked here ($u$ is a $C^2$ solution of $u_t - \Delta u = f(u)$ and $u=0$ on $\partial\Omega \times (0,\infty)$. Show if $u(x,0) \geq 0$, then $u(x,t) \geq 0$)
However, my question is if $u(x,0)\leq C$ for all $x \in \Omega$, then how to show that $u(x,t)\leq Ce^{Mt}$ for all $x \in \Omega$ and $t>0$?
Here's an idea you can try. Let $v(x,t)$ solve $\partial_tv-\Delta v=f(v)$ for $v(x,0)=C$, and $w=v-u$. Then $w$ solves $\partial_tw-\Delta w=f(v)-f(u)$. By the fundamental theorem of calculus, $$ f(v)-f(u)=(v-u)\int_0^1 f'((1-s)u+sv)ds=:w\,F, $$ where $F(x,t)=\int_0^1f'((1-s)u(x,t)+sv(x,t))ds$. So $w$ solves $$ \partial_tw-\Delta w=wF. $$ But $wF=0$ when $w=0$, and $w(x,0)\ge 0$, so the supersolution argument you linked to seems to apply. Note that $v(x,t)\le Ce^{Mt}$, which is straightforward to show, since $v(x,t)=V(t)$ solves $V'(t)=f(V(t))$.