I think the 'identical equation' Sylvester has in mind is

$$\sum_q \mu(q) \frac{x^q}{1-x^{2q}} = x+x^5+x^{13}+x^{17}+x^{25}+x^{29}+\cdots $$

where the left-hand sum is over all natural numbers $q$ divisible only by primes of the form $4s+3$ and the right hand side is the sum of all powers $x^r$ where $r$ is divisible only by primes of the form $4s+1$. (Sylvester specifies no repeated prime factors for $q$ on the left-hand side, but since I'm using $\mu$, any such summand is killed by $\mu(q) = 0$; note the first summand is for $q=1$.)

*Proof.* The left coefficient of $x^n$ in the left-hand side is $\sum_{q} \mu(q)$ where the sum is over all square-free $q$ divisible only by primes of the form $4s+3$ such that $n/q$ is odd. It is therefore zero for even $n$. If $n$ is odd let $n = Np_1\ldots p_t$ where $p_i \equiv 3$ mod $4$ for each $i$ and no such prime divides $N$. The sum is then

$$\sum_{q \mid p_1\ldots p_t} \mu(q) = \begin{cases} 1 & \text{if $t=0$} \\ 0 & \text{otherwise.} \end{cases} $$

Hence the coefficient of $x^n$ is $0$ unless $n$ is divisible only by primes of the form $4s+1$, in which case it is $1$. $\Box$

The rest of Sylvester's argument seems clear enough to me: if there are only finitely many primes of the form $4s+3$ then the left-hand side is a finite sum, and well-defined when $x=i$ since $i^{2q} = (-1)^q = -1$ as $q$ is odd, and so $1-x^{2q} = 2$. But then, by hypothesis (and infinitely many primes), there are infinitely many primes of the form $4s+1$, making the right-hand side infinite when $x=i$.

There is of course an easier argument, as Euclid take the product of finitely many primes of the form $4s+3$ including the prime $3$, multiply by $4$ and subtract $1$; the result is then divisible by another prime still of the form $4s+3$.

Since I had to make an edit anyway, I'll add that almost the same argument works for primes of the form $6s+1$ and $6s+5$; using the latter instead of primes of the form $4s+3$ to define the left-hand side, the right-hand side is the sum of all powers $x^r$ where $r$ is divisible only by the prime $3$ or primes of the form $6s+1$. But again one can show there are infinitely many primes of the form $6s+5$ by a variation on Euclid's argument.

One feature of interest to me is that Sylvester's argument uses Lambert series rather than the Dirichlet series ubiquitous in analytic number theory.