Not sure how much about complex numbers you know so bear with me. I don't like faffing around with multivalued stuff when I can help it, everything here is single-valued.
The reasoning in the video is incorrect and the comments point this out. So let's do this all properly. For $a \in \mathbb C \setminus \{0\}$ and $z \in \mathbb C$, we should define $a^z = \exp(z (\ln|a| + i \arg(a))$ for some choice of argument $\arg$ and where $\ln$ is the real logarithm. These choices of argument all differ by $2 \pi$ (because $\exp$ is periodic in $2 \pi i$). An "argument" here is a real number $\theta$ such that $z = |z| e^{i \theta}$.
If we choose $\arg 1 = 0$, we have $1^z = \exp(z \ln 1) = \exp(z \cdot 0) = 1$ for all $z \in \mathbb C$ and there are therefore no solutions.
If we instead choose $\arg 1 = 2 \pi$, we instead have $1^z = \exp(z (\log 1 + 2 \pi i)) = \exp(2 \pi i z)$. So we have the equation $\exp(2 \pi i z) = 2$. Now, we know that $\exp(z) = 1$ if and only if $z = 2 \pi i n$ for some integer $n$. We can show this by writing out $z = x + i y$ and writing $$\exp(z) = \exp(x + i y) = \exp(x) \exp(i y) = \exp(x)(\cos y + i \sin y) = 1$$ and comparing real and imaginary parts. Since $\exp(x - y) = \exp(x)\exp(-y)$, this means that if $\exp(z) = \exp(w)$, we have $\exp(z - w) = 1$ and so $z = w + 2 \pi i n$ for some integer $n$. So in the equation $\exp(2 \pi i z) = 2 = \exp(\ln(2))$, we obtain $2 \pi i z = \ln(2) + 2 \pi i n$, and so $z = n + \frac {\ln(2)} {2 \pi i} = n - \frac {i \ln(2)} {2 \pi}$ for integer $n$.
So where does this $-\frac {i \ln(2)} {2 \pi n}$ come from in the single-valued perspective? Well, if we in general pick $\arg 1 = 2 \pi k$ for some integer $k$, we will get that the solutions are $z = n - \frac {i \ln(2)} {2 \pi k}$ for integer $n$. So $-\frac {i \ln(2)} {2 \pi n}$ is a solution if you pick $\arg(z) = 2 \pi n$. They do however miss infinitely many additional solutions by not adding on integers.
If you are to mess around with multi-valued functions, we basically define $\log$ to be a set rather than a number which captures all the possible values of $\log$, ie. all the possible choices of $\arg$. So you get $\log z = \{\ln|z| + i\theta : \theta \text { satisfies } |z|e^{i \theta} = z\}$ say. So $\log 1 = \{\ldots, -2 \pi i, 0, 2 \pi i, \ldots\} = \{2 \pi i n : n \in \mathbb Z\}$. I guess you then define $1^z = \exp(z \log 1) = \{\exp(2 \pi i n z) : n \in \mathbb Z\}$, technically the image of $\exp$ under the set $z \log 1 = \{\ldots, -2\pi i z, 0, 2 \pi i z\}$. The question of whether $1^z = 2$ is then to be interpreted as which $z \in \mathbb C$ satisfy $2 \in \{\exp(2 \pi i n z) : n \in \mathbb Z\}$, in which case the answer is $z_{n, k} = n - \frac {i \ln(2)} {2 \pi k}$ for any integer $n$ and any integer $k$. Specifically, you will have $\exp(2 \pi i k z_{n, k}) = 2$. This is a very awkward sense of a "solution" though, in my opinion (maybe it's just me), but it more or less formalises what is being spoken about here.
