For the first part:
$$100a+10b+c = a+b^2+c^3$$
implies that $$99a + b(10-b)=(c-1)c(c+1)$$
Now, $b(10-b)\leq 25$, with $b(10-b)=0,9,16,21,24,25$. So you want $c$ so that $$(c-1)c(c+1)\equiv 0,9,16,21,24,25\pmod{99}.$$
Now, $(c-1)c(c+1)$ is divisible by $3$, as is $99$, so you really need $$c^3-c\equiv 0,9,21,24\pmod {99}, \,c^3-c\geq 100$$.
This can be done by trial and error.
We know $c>4$. (If $c\leq 4$ then we get $a=0$, which you might want to include.)
$c=5$ gives $c^3-c\equiv 21\pmod {99}$.
So $\overline{abc}=135, 175$ are solutions.
$6^3-6\equiv 12\pmod{99}$, so $c\neq 9$.
$7^3-7\equiv 39\pmod{99}$, so $c\neq 7$.
$8^3-8\equiv 9\pmod{99}$ so $(a,b,c)=(5,1,8)$ and $(a,b,c)=(5,9,8)$ are solutions.
$9^3-9\equiv 27\pmod{99}$. So $c\neq 9$.
This yields the four solutions: $135, 175, 518, 598$.
(Allowing $a=0$, we get additional answers $\overline{abc}=000,001,043,063$.)
A start on the bonus question.
For $n$ digits, the largest sum $a_1+a_2^2+\cdots a_2^n$ is $9+9^2+\cdots +9^n = \frac{9}{8}(9^n-1)$. The smallest $n$-digit number is $10^{n-1}$.
So if $10^{n-1}>\frac{9}{8}(9^n-1)$, you can be sure there is no solution. But $$\frac{9}{8}(9^n-1)< \frac{9^{n+1}}{8},$$ so if $10^{n-1}>9^{n+1}/8,$ there is no solution. This is true if $(n-1)\log_9(10)>(n+1)-\log_9 8$ or $$\log_9(10) > 1+\frac{2-\log_9 8}{n-1}\\\log_9(10/9)>\frac {2-\log_9 8}{n-1}$$ or $$n>1+(2-\log_9(8))\log_{10/9}(9)\approx 22.97$$
So there are no solutions with more than 22 digits. That limits it to a finite, if large, problem.
Best Answer
The tens digit of $n$ can be read off of $n\pmod {100}$. In this case, $7^4\equiv 1 \pmod {100}$ so $7^{100}\equiv 1 \pmod {100}$, so the answer is $0$.