$a^p \equiv a \pmod{p}$ by Fermat's little theorem but by hypothesis $a^q \equiv a\pmod{p}$, therefore
$$a^{pq} \equiv (a^p)^q \equiv a^p \equiv a \pmod{p}$$.
Do the same with the other one, and then use the CRT.
EDIT:
By doing the same to the other one you'll get:
$$a^{pq} \equiv (a^q)^p \equiv a^q \equiv a \pmod{q} $$
Now the CRT is useful here only when you want to conclude that $a^{pq} \equiv a \pmod{pq}$.
This is what the Chinese Remainder theorem says for the case of two congruence systems:
If we're looking for $x$ such that $x \equiv a \pmod{p}$ and $x \equiv b \pmod{q}$ for two distinct primes (or two coprime moduli) then the CRT asserts that such an $x$ exists and it is unique $\mod {pq}$.
Now in our case, you can use the CRT by thinking that both $a^{pq}$ and $a$ satisfy $x \equiv a \pmod{p}$ and $x \equiv a \pmod{q}$. So, by applying the uniqueness part of the CRT, we must have $a^{pq} \equiv a \pmod{pq}$.
I don't like this approach though. All you need to know is that the least common multiple of two relatively prime numbers is their product, then by using the definition of the least common multiple, you can say that if $p \mid a^{pq}-a$ and $q \mid a^{pq}-a$ then $pq \mid a^{pq}-a$ because $\operatorname{LCM}(p,q)=pq$. This one is a better approach I think.
If $x\equiv 1\pmod 7$ and $x\equiv 1\pmod {13}$ then $x\equiv 1\pmod {91}$.
If $x\equiv -1\pmod 7$ and $x\equiv -1\pmod {13}$ then $x\equiv -1\pmod {91}$.
If $x\equiv 1\pmod 7$ and $x\equiv -1\pmod {13}$ then $x\equiv 64\pmod {91}$.
If $x\equiv -1\pmod 7$ and $x\equiv 1\pmod {13}$ then $x\equiv 27\pmod {91}$.
In each case, the chinese remainder theorem guarantees that the solution you found (by trial-and-error) $\pmod{91}$ is the only one.
Best Answer
As explained in the comments, it didn't work because CRT wasn't applied correctly.
In particular, the solution to $ 2^k + k \equiv 0 \pmod{17}$ has a cycle length of $ 16 \times 17$, because $2^k$ has a cycle length of 16 and $k$ has a cycle length of 17.
So, the (theoretical) approach is to find all the solutions to $2^k+k \equiv 0 \pmod{17} $ working in $\pmod{17 \times 16}$, and likewise for the other equation working in $\pmod{19 \times 18}$, and finally combine them via CRT in $ \pmod{17 \times 19 \times 144}$.
If you understand the above logic, I do not recommend actually finding all the solutions, as it's just a very tedious process that's best left to the computer.