The discriminant form $\Delta$ is essentially unique

First, holomorphy and weight: $E_4,E_6$ are modular forms of weight $4,6$ respectively (standard Eisenstein-series facts, taken as known input here), so $E_4^3$ and $E_6^2$ are both weight $12$. Both have Fourier expansion starting with constant term $1$ (by normalization), so $E_4^3-E_6^2$ has vanishing constant term: it is a cusp form of weight $12$, and $\Delta=(E_4^3-E_6^2)/1728$ is too, in particular $\operatorname{ord}_\infty(\Delta)\ge1$.

Now apply the valence formula (proved above) with $k=12$: the total weighted zero order of $\Delta$ must equal $12/12=1$. Every term on the left — $\operatorname{ord}_\infty(\Delta)$, $\tfrac12\operatorname{ord}_i(\Delta)$, $\tfrac13\operatorname{ord}_\rho(\Delta)$, and any $\operatorname{ord}_P(\Delta)$ for other points — is a non-negative real number, and we already know $\operatorname{ord}_\infty(\Delta)\ge1$. The only way a sum of non-negative terms, one of which is already $\ge1$, can total exactly $1$, is if that term equals exactly $1$ and every other term is exactly $0$. Hence $\operatorname{ord}_\infty(\Delta)=1$ exactly, and $\operatorname{ord}_P(\Delta)=0$ for every $P\in\mathbb{H}$: $\Delta$ never vanishes on $\mathbb{H}$, with a simple zero only at the cusp.

For the dimension claim: let $g$ be any weight-$12$ cusp form. The quotient $g/\Delta$ is $\mathrm{SL}_2(\mathbb{Z})$-invariant (weight $12-12=0$), and holomorphic on $\mathbb{H}$ since $\Delta$ never vanishes there. At the cusp, both $g$ and $\Delta$ vanish to order at least $1$ in $q$, and since $\Delta$'s order there is exactly $1$, the quotient extends holomorphically (no pole) to $q=0$ as well. A weight-$0$ modular function holomorphic everywhere on $\mathbb{H}$ and at the cusp descends to a holomorphic function on the compact Riemann surface $X(1)\cong\mathbb{P}^1(\mathbb{C})$, which by Liouville's theorem (extended to the compactification) must be constant. So $g=c\cdot\Delta$ for a constant $c$: the space of weight-$12$ cusp forms is exactly $1$-dimensional. The identical argument at weight $4$ (total budget $4/12=1/3$, which can only be realized as $\operatorname{ord}_\rho=1$ with every other order $0$, since any integer contribution — such as a nonzero $\operatorname{ord}_\infty$ — would already exceed $1/3$) shows every nonzero weight-$4$ form has its unique zero at $\rho$ with order exactly $1$ and no others, and the same quotient-by-$E_4$ trick forces $\dim M_4(\mathrm{SL}_2(\mathbb{Z}))=1$.

Finally, the $q$-product $\Delta(\tau)=q\prod_{n\ge1}(1-q^n)^{24}$ is the classical identity discovered by Jacobi; we cite it here rather than re-derive it, but note it is fully consistent with what we just proved: the right-hand side visibly vanishes to order exactly $1$ at $q=0$ and is manifestly nonzero for $0<|q|<1$ (a convergent product of nonzero factors), matching the zero-order profile forced by the valence formula above.