State of knowledge of $a^n+b^n=c^n+d^n$ vs. $a^n+b^n+c^n=d^n+e^n+f^n$ As far as I understand, both of the Diophantine equations
$$a^5 + b^5 = c^5 + d^5$$
and
$$a^6 + b^6 = c^6 + d^6$$
have no known nontrivial solutions, but
$$24^5 + 28^5 + 67^5 = 3^5+64^5+62^5$$
and
$$3^6+19^6+22^6 = 10^6+15^6+23^6$$
among many other solutions are known, when the number of summands is increased
from $2$ to $3$.
My information here is at least a decade out of date,
and I am wondering if the resolution of Fermat's Last Theorem has clarified
this situation,
with respect to sums of an equal number of powers...?
 A: The short answer to your specific question is no, the resolution of FLT via modularity of elliptic curves does not seem to be helpful in dealing with rational points on higher dimensional varieties. The first two equations you list, $a^5+b^5=c^5+d^5$ and $a^6+b^6=c^6+d^6$, are surfaces of general type in $\mathbb{P}^3$, so conjectures of Bombieri, Lang, Vojta, would say that their solutions, and solutions of similar equations in which the terms might have non-unit coefficients, lie on finitely many curves. The second two equations are 4-folds in $\mathbb{P}^5$. For $x^5+y^5+z^5=u^5+v^5+w^5$, the canonical bundle is ample, and I'm going to guess that the 4-fold is rational(?), so there will be lots of solutions. For $x^6+y^6+z^6=u^6+v^6+w^6$, the canonical bundle is trivial, so you've got a Calabi-Yau. At least conjecturally, there should be a number field $K$ such that the $K$-rational points are Zariski dense. Maybe for this 4-fold, one can take $K=\mathbb{Q}$?
A: Label the equation,
$$x_1^k+x_2^k+\dots+x_m^k = y_1^k+y_2^k+\dots+y_n^k$$
as a $(k,m,n)$. Let type of primitive solutions be polynomial identity $P(n)$, or elliptic curve $E$. Then results (mostly) for the balanced case $m=n$ are,
I. Table 1
$$\begin{array}{|c|c|c|}
(k,m,n)& \text{# of known solutions}& \text{Type}\\
3,2,2& \infty&P(n)\\
4,2,2& \infty&P(n),E\\
5,3,3& \infty&P(n),E\\
6,3,3& \infty&P(n),E\\
7,4,4& \text{many} &E\,?\\
7,4,5& \infty&P(n)+E\\
8,4,4& 1&-\\
9,5,5& \text{many}&-\\
9,6,6& \infty & E\\
10,5,5& 0&-\\
\end{array}$$
Note: For $(7,4,5)$, see this MSE answer. 
II. Table 2. (For multi-grades)
$$\begin{array}{|c|c|c|}
(k,m,n)& \text{# of known solutions}&\text{Type}\\
5,4,4& \infty&P(n),E\\
6,4,4& \infty&P(n),E\\
7,5,5& \infty&P(n),E\\
8,5,5& \infty&E\\
9,6,6& \text{many}&E\,?\\
10,6,6& \infty&E\\
11,7,7& 0&-\\
12,7,7& 0&-\\
\end{array}$$
Note: A multi-grade is simultaneously valid for multiple $k$. For example, the $(9,6,6)$ is for  $k=1,3,5,7,9$ while the $(10,6,6)$ is for $k=2,4,6,8,10$. 
