As far as I know there are two approaches to Goldbach type problems, the circle method and sieve methods. In the sequel I will restrict myself to the circle method, hoping that someone else writes something about sieves.
Define the exponential sum $S(\alpha)=\sum_{n\leq N}\Lambda(n) e(n\alpha)$, where $e(x)=e^{2\pi i x}$. By orthogonality of $e(x)$ we have that the (weighted) number of representations of $N$ as the sum of 3 primes is $$ \int_0^1 S(\alpha)^3e(-N\alpha)\;d\alpha. $$ If $\alpha=\frac{p}{q}$ is rational, then $$ S(\alpha)=\sum_{(a,q)=1} e(\frac{ap}{q})\underset{n\equiv a\pmod{q}}{\sum_{n\leq N}} \Lambda(n), $$$$ S(\alpha)=\sum_{a=0} ^{q-1} e \left( \frac{ap}{q} \right) \underset{n\equiv a\pmod{q}}{\sum_{n\leq N}} \Lambda(n), $$ which can be approximated using the prime number theorem for arithmetic progressions. If $\alpha$ is very close to a rational number, evaluation can still be done by partial summation. Hence define the major arcs $\mathfrak{M}\subseteq[0,1]$ as the set of integers very close to rational numbers with small denominator. Then $$ \int_\mathfrak{M}S(\alpha)^3e(-N\alpha)\;d\alpha $$ can approximately be evaluated, it turns out that this integral essentially equals the expected main term. Let $\mathfrak{m}=[0,1]\setminus\mathfrak{M}$ be the so called minor arcs. To prove that every sufficiently large odd integer is the sum of three primes, Vinogradoff showed that for suitably defined arcs the integral over the major arcs is $\sim\mathfrak{S}(N)N^2$, and that for $\alpha\in\mathfrak{m}$ we have $|S(\alpha)|\ll\frac{N}{\log^2 N}$. Hence \begin{eqnarray*} \int_\mathfrak{m}S(\alpha)^3e(-N\alpha)\;d\alpha & \leq &\int_\mathfrak{m}|S(\alpha)^2|\;d\alpha\max_{\alpha\in\mathfrak{m}}|S(\alpha)|\\ & \leq & \int_0^1|S(\alpha)^2|\;d\alpha\max_{\alpha\in\mathfrak{m}}|S(\alpha)|\\ & = & \sum_{n\leq N}\Lambda(n)^2\max_{\alpha\in\mathfrak{m}}|S(\alpha)|\\ & \ll & \frac{N^2}{\log N} \end{eqnarray*} So for large $N$ the error coming form $\mathfrak{m}$ is of smaller order then the main term, and we obtain an asymptotic for the number of representations. Since proper powers are pretty rare, passing from $\Lambda$ to primes is no problem.
All this was known and used by Hardy and Littlewood some 20 years before Vinogradov, however, they could not give an unconditional bound for $S(\alpha)$, and had to assume the generalized Riemann hypothesis. Among the things Helfgott did, was a numerical bound which is non-trivial for rather small values of $N$, which in analytic number theory is usually pretty difficult. However, no matter what progress may come, the argument used for odd $N$ can never prove binary Goldbach. The crucial point in the argument above was that the integral over the major arcs is asymptotically equal to the conjectural main term, and the integral over the minor arcs is of smaller order. The integral over the major arcs can still be evaluated, and is $\mathfrak{S}(N)N$, as it should, however, $$ \int_0^1 |S(\alpha)|^2\;d\alpha = \sum_{n\leq N}\Lambda(n)^2\sim N\log N $$ is bigger then the main term. To prove Goldbach for even integers we have to show that $\left|\int_\mathfrak{m} S(\alpha)^2e(-N\alpha)\;d\alpha\right|$ is considerably smaller than $\int_\mathfrak{m} |S(\alpha^2)|\;d\alpha$, that is, that there is some cancelation within the integral.
Still, one can obtain interesting results in this way. For example, Montgomery and Vaughan (The exceptional set in Goldbach's problem, Acta Arith. 27) have shown that all even integers up to $x$ with at most $x^{1-c}$ exceptions can be written as the sum of two primes. They used the fact that $\int_\mathfrak{m} S(\alpha)^2e(-N\alpha)\;d\alpha$ cannot simultaneously be large for many different $N$.
Somewhat surprisingly Brüdern, Granville, Perelli, Vaughan, and Wooley, (On the exponential sum over k-free numbers, R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci. 356) showed that there are interesting binary additive problems which can be solved by the circle method, e.g. problems involving squarefree integers. However, as they explicitly mention, their method fails for sets of density 0, the reason being the same as above.