What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \\ a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \\ \vdots $$ we instead considered $$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \\ a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \\ a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \\ \vdots $$ My gut feeling is that this will have infinitely many solutions.
Note: $a_i,b_i$ are all positive integers.
Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, z\in \mathbb{Z}$ with $z\neq \pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as $$ (z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3} $$ for $m\geq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions. If you only want positive solutions, it is also possible because we have $$ 1214928^{3} + 3480205^{3} = 3528875^{3} + 2. $$ Similarly, it is enough to show that there exists an integer $1\leq t \leq k-1$ such that the equation $$ x_{1}^{k} + x_{2}^{k} + \cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k $$ has an integer (or positive integer) solution with $y\neq \pm 1$, which will give a parametrization $$ (y^{m})^{k} + \cdots + (y^{m})^{k} + (y^{m+1})^{k} + \cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + \cdots + (x_{k-1}y^{m})^{k} $$ where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.