Given indeterminates $x$ and $y$ and a field $k,$ consider the polynomial ring $R = k[x^4, x^3 y, x^2 y^2, xy^3, y^4] \subseteq k[x, y].$ Observe that the irrelevant maximal ideal of $R$ is given by $\mathfrak m = (x^4, x^3 y, x^2 y^2, xy^3, y^4),$ and its square is simply $\mathfrak m^2 = (x^i y^{8 - i} \,|\, 0 \leq i \leq 8).$ Consider a set $S \subseteq \{0, 1, 2, \dots, 8\}$ such that the ideal $I = (x^s y^{8 - s} \,|\, s \in S)$ satisfies $I = \mathfrak m^2.$
Claim: we must have that $S = \{0, 1, 2, \dots, 8\}.$
Considering that $\mathfrak m^2 = I,$ it follows that $I$ contains each generator of $\mathfrak m^2.$ Consequently, for any integer $0 \leq i \leq 8,$ we may write $x^i y^{8 - i}$ as an $R$-linear combination of the generators of $I$, i.e., $$x^i y^{8 - i} = \sum_{s \in S} f_s x^s y^{8 - s} \text{ for some polynomials } f_k \in R.$$ From here, I would like to make a degree argument to show that $i = s,$ and therefore, we have that $\{0, 1, 2, \dots, 8\} \subseteq S$; however, this all feels very ad hoc, and I cannot help but wonder if there is a more elegant (or correct) argument to be made.
If I were to make a degree argument, I imagine that it would go something along the lines of the following. Comparing the multidegrees of the left- and right-hand sides, we must have that $$x^i y^{8 - i} = \sum_{s \in S} g_s x^s y^{8 - s}$$ for some monomials $g_s$ of multidegree $(i - s, s - i).$ But as $g_s$ are monomials in $k[x, y],$ it follows that $i - s \geq 0$ and $s - i \geq 0$ so that $i = s.$ Consequently, we have that $\{0, 1, 2, \dots, 8\} \subseteq S.$
I would be extremely grateful for some guidance on how to approach this problem.