Let $G=(g_1,\ldots,g_r)$ and $G'=(g'_1,\ldots,g'_s)$ be two minimal Gröbner bases for $I$ (w.r.t. the same monomial ordering).
Applying the fact that $G'$ is a Gröbner basis to $LT(g_i)$ yields $g'_k$ such that $LT(g'_k) \mid LT(g_i)$. Conversely, applying the fact that $G$ is a Gröbner basis to $LT(g'_k)$ yields $g_\ell$ such that $LT(g_\ell) \mid LT(g'_k)$. Putting the two together gives $LT(g_\ell) \mid LT(g_i)$. Since $LT(g_i)\not\in\langle LT(G-\{g_i\})\rangle$ by minimality of $G$, it follows that $\ell = i$. Now we have $LT(g_i) = cLT(g'_k)$ for some constant $c \neq 0$. From the minimality of both $G$ and $G'$ it follows that $1 = LC(g_i) = cLC(g'_k) = c$, so $LT(g_i) = LT(g'_k)\in LT(G')$.
The simplest systems of polynomial equations are the ones that can be solved using back-substitution. Here is one way to contruct such a system with exactly $m \geq 1$ common solutions: pick your favorite polynomial
\begin{equation*}
f(y) = (y-\alpha_1)\cdot\dots\cdot(y-\alpha_{m-1})
\end{equation*}
with $\alpha_1,\dots,\alpha_{m-1} \neq 0$ all distinct and consider the system $V(G)$, where $G = \{x,yf(y)\}$. It follows from Buchberger's criterion that $G$ is a Gröbner basis since $S(x,yf(y))$ is divisible by $x$. Therefore, $I = \langle x,yf(y) \rangle$ is an ideal in $K[x,y]$ with Gröbner basis $G$ such that $V(G) = \{(0,0),(0,\alpha_1),\dots,(0,\alpha_{m-1})\}$ has exactly $m$ solutions. The only problem is that $G$ has $2$ elements instead of $3$, but this can be easily remedied: just add any other polynomial in $I$ to $G$, e.g. consider $G' = \{x,yf(y),xy\}$. Then $G'$ is a Gröbner basis for $I$ consisting of $3$ polynomials with exactly $m$ common solutions, as desired.
There is still something unsatisfactory with the solution presented above: the Gröbner basis $G'$ is not reduced. In other words, we would like to answer the question:
Find an ideal $I \subseteq K[x,y]$ whose reduced Gröbner basis with respect to the graded lexicographic order (which is unique) consists of 3 polynomials with exactly $m$ common solutions.
One idea to achieve this is to enlarge $G$ by adding a polynomial that is not in $I$ but that vanishes at all the $m$ common solutions of $G$. Actually, it helps to first modify $G$ to $\{x^2,y^2f(y)\}$. We may then consider the basis $G' = \{x^2,xy,y^2f(y)\}$ and the ideal $I = \langle x^2,xy,y^2f(y) \rangle$ it generates. Since $S(x^2,xy) = 0$, $S(x^2,y^2f(y))$ is divisible by $x^2$ and $S(xy,y^2f(y))$ is divisible by $xy$, it follows from Buchberger's criterion that $G'$ is a Gröbner basis of $I$. Moreover, note that $G'$ is reduced and consists of $3$ polynomials with exactly $m$ common solutions.
Finally, the same argument can be used to prove that the set
\begin{equation*}
\{x^{n-1},x^{n-2}y,\dots,xy^{n-2},y^{n-1}f(y)\}
\end{equation*}
is a reduced Gröbner basis for the ideal it generates which consists of $n \geq 2$ polynomials with exactly $m \geq 1$ common solutions.
Best Answer
The set $\mathbb{M}$ of all monomials forms a semigroup, and the set of leading monomials $LM(I)$ is a semigroup ideal in $\mathbb{M}$. The ideal basis $G=(g_1,\ldots,g_s)$ is a Gröbner basis of $I=\langle g_1,\ldots,g_s\rangle$ iff $LM(I)$ is generated as a semigroup ideal by $LM(g_1), \ldots, LM(g_s)$.
The two definitions are equivalent.