= Solution
Put $I=(f_1,\ldots,f_r)$ and $\mathfrak m=(y_0,\ldots,y_n)$. If $\mathfrak m^N\subseteq I$, a projective common zero would have some coordinate $y_j\ne0$. For $N>0$, the polynomial $y_j^N\in I$ could not vanish there; for $N=0$, the inclusion puts $1$ in $I$, also impossible. Thus the projective zero set is empty.
Conversely, over an <algebraically closed field>, emptiness of the projective zero set means the affine common zero set is contained in the origin. Every $y_j$ vanishes there, so the <Hilbert Nullstellensatz> puts $y_j$ in $\sqrt I$. Choose integers $e_j\ge1$ with $y_j^{e_j}\in I$, and set
$$
N=1+\sum_{j=0}^n(e_j-1).
$$
Every degree-$N$ <monomial> has some exponent at least the corresponding $e_j$, by the pigeonhole principle, and is consequently in $I$. Those <monomials> generate $\mathfrak m^N$, proving
$$
\boxed{V_+(I)=\varnothing\iff\mathfrak m^N\subseteq I\text{ for some }N\ge0.}
$$
The argument includes the unit-ideal case. This proves the <Projective Nullstellensatz> criterion directly from its affine counterpart.
Over a general field, this same criterion concerns geometric zeros: apply the proof after extension to an algebraic closure. Inclusion of each degree-$N$ <monomial> descends back to $k$, since its class in $k[y]/I$ is zero if and only if it becomes zero after a field extension; tensoring a <vector space> with an extension field is faithful. If instead one counts only $k$-rational projective points, the assertion is false. For example $y_0^2+y_1^2$ has no real projective zero but has the complex zero $[1:i]$, so its <ideal> cannot contain any power of $\mathfrak m$.
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