Let $P_k(F)$ denote the $F$-vector space of (univariate) polynomials of degree $\leq n$. Letting $F$ be a field lets everything be monic, but it seems sufficient to consider a ring $R$ such that the coefficients of the leading terms of $f,g$ aren't zero-divisors.
Let $F$ be a field. Consider the linear map $$ T_{f,g}: P_{n-1}(F) \oplus P_{m-1}(F) \to P_{m+n-1} \; s.t. \; T_{f,g}(a,b) = af + bg$$ where $f,g$ are polynomials such that $\deg f = n$ and $\deg g = m$. It is claimed that $$\dim \left(\operatorname{img}(T_{f,g})\right) = m+ n - d$$ where $d = \deg \gcd[f,g]$. By rank-nullity this claim can be stated as $$ \deg \gcd[f,g] = \dim \ker T_{f,g}$$
Note that the matrix corresponding $T_{f,g}$ is in fact the transpose of the Sylvester matrix. As rank is invariant under transposition the claim can be equivalently stated in a third way as $$\operatorname{rk}(\operatorname{Syl}(f,g)) = m+ n - d $$
My question is how to prove this. The second formulation seems the most accessible, and roughly speaking the task is to show that the solutions $(a,b)$ of $$ af + bg = 0$$ form a subspace of dimension $d$. Bezout's identity for integers appears to be related, at least insofar as the Wikipedia page uses it to handwave the proof.
Preferably I would like a solution that does not use the fact that $\det T_{f,g}$ is in fact the resultant. This is my current attempt:
Let $h = \gcd[f,g]$ and $ab+fg = 0$. Writing $f = h f'$, $g = hg'$ we have
$$af+bg = h(af' +b g') = 0 \implies af' + bg' = 0 \implies af' = -bg'$$ and by the definition of $h$, $f' \nmid g$, $g' \nmid f$, giving $$ f' | -bg' \implies f' | -b \:, \:g' | af' \implies g' | a$$ and therefore $a = k g'$, $b = -k f'$ for $(k,k) \in P_{n-1}(F) \oplus P_{m-1}(F)$. Now, as $f= hf'$, $g=hg'$ we have $\deg f' = m-d$, $\deg g' = n-d$.
Finally as $(kg',-kf') \in P_{n-1}(F) \oplus P_{m-1}(F)$, we have $\deg k \leq d-1$ and consequently $$\ker T_{f,g} = F\{(kt^i,kt^i): 0 \leq i \leq d-1 \}$$ and clearly $\dim \ker T_{f,g} = d$.