I think there is an elementary proof, but let's check it.
I'm relying on:
There always exist maximal proper homogeneous right ideals (by the usual Zorn's Lemma argument.)
The sum of two homogeneous right ideals is again homogeneous.
Suppose $M$ and $N$ are distinct maximal homogeneous right ideals. Then $M+N=R$, and there exists $m+n=1$ with $m\in M$ and $n\in N$. Because of the grading, the grade zero parts must be such that $m_0+n_0=1$, and because $M$ and $N$ are both proper and homogeneous, neither $m_0$ nor $n_0$ can be units of $R_0$. This implies $R_0$ is not local.
By contrapositive then, we have shown if $R_0$ is local, then $R$ is graded local.
After some research, I think I can answer this question myself.
If $R$ is $\mathbb{N}$-graded over a field $K$ with homogeneous maximal ideal $R^+$ and $I$ is a homogeneous ideal, it will always have a minimal homogeneous generating set. The idea is that $I/(R^+)I$ is a graded $K$-vector space, and has a homogeneous basis consisting of $\mu_h(I)$ elements.
But, if we consider instead the local ring $(R_{R^+},(R^+)R_{R^+},K)=(S,\mathfrak{m},K)$ and let $J=IS$, $J/\mathfrak{m}J$ is canonically isomorphic to $I/(R^+)I$ as $K$-vector spaces, since $I$ and $R^+I$ are inside $R^+$. Then, $\mu(J)=\mu(I)$ and $\mu(J)=\dim_K(J/\mathfrak{m}J) = \dim_K(I/(R^+)I)=\mu_h(I)$. From this, we can even get the number of minimal generators of $I$ of degree $m$ as the dimension of the $K$-vector space $[I/(R^+)I]_m$. This is a standard argument about graded Betti numbers as well, see this MathOverflow answer.
The reason this proof doesn't apply to the example in the answer to this previously linked question is that the $\mathbb{Z}$-graded ring mentioned does not contain a field in its degree 0 part.
This also settles the existence of homogeneous reductions of the maximal ideal. In fact, a quick graded adaptation of the proof of the existence of minimal reductions in the excellent paper "Reductions of Ideals in Local Rings" by D. Northcott and D. Rees will guarantee homogeneous minimal reductions also exist. It does seem possible that the reduction number measured using only homogeneous minimal reductions may be larger than the reduction number measured using all minimal reductions, but I am not sure about this.
Best Answer
There are two questions here:
1) One is asking if $\cdots\oplus R_{-1}\oplus m_0\oplus R_1\oplus\cdots$ is an ideal in a $\mathbb Z$-graded ring $R$, where $m_0$ is a maximal ideal of $R_0$, and the answer is negative as shows the following example: $R=k[t,t^{-1}]$, and (necessarily) $m_0=(0)$.
2) The other is an exercise in Marley's notes asking to prove that any homogeneous and maximal ideal $M$ in a $\mathbb Z$-graded ring $R$ has the form $M=\cdots\oplus R_{-1}\oplus m_0\oplus R_1\oplus\cdots$, where $m_0$ is a maximal ideal of $R_0$. In this case $R/M$ is a graded ring and also a field. This shows that $(R/M)_n=0$ for all $n\ne 0$ (why?), that is, $R_n=M\cap R_n$ for all $n\ne 0$. If $m_0=M\cap R_0$ it follows that $M=\cdots\oplus R_{-1}\oplus m_0\oplus R_1\oplus\cdots$.