Verifying that $2^{44}-1$ is divisible by $89$

As the title, I was asked to show that $2^{44}-1$ is divisible by $89$.

My first attempt is to change the expression into $(2^{22}-1)(2^{22}+1)$.

Then further simplified it into $(2^{11}-1)(2^{11}+1)(2^{22}+1)$, I used my calculator and was able to show that $2^{11}-1$ is divisible by $89$ but then I don't know how to show it with modular arithmetic. I do think that it is quite similar to the form where we can use the Fermat's little theorem. $(\sqrt{2})^{88}-1$. (Though I do understand Flt can only be applied to integer.)

Can someone tell me whether I can take square root in modular arithmetic as well? I am still pretty new to the modular arithmetic. Thank you very much.

Hint: By Fermat's Theorem, we have $2^{88}\equiv 1\pmod{89}$.

So $(2^{44}-1)(2^{44}+1)\equiv 0 \pmod{89}$.

If we can show that $2^{44}+1\not\equiv 0\pmod{89}$ we will be finished.

One way to do this is to use the fact that $2$ is a quadratic residue of $89$, since $89$ is of the shape $8k+1$.

Remark: Your direct computational approach is perfectly fine. However, it may be that you are expected to make use of "theory," as in the approach described above.

An idea:

$$2^5=32=-57\pmod{89}\;\;,\;\;2^6=64=-25\pmod{89}\implies$$

$$2^{11}=32(-25)=-800=1\pmod{89}$$

since $\;801=9\cdot 89\;$