$\alpha^{2^{m} + 1}$ is a generator of $\operatorname{GF}(2^m)$? This is from a paper (Partitions in the S-Box of Streebog and
Kuznyechik) about S-Boxes:

Let $\operatorname{GF}(2^{2m}) = \mathbb{F}_2[X]/p(X)$ be a finite field of even degree defined by a primitive polynomial $p$. The multiplicative subgroup $\operatorname{GF}(2^{2m})^*$ is cyclic and generated by $\alpha$ which is such that $p(\alpha) = 0$.
In this context, $\alpha^{2^{m} + 1}$ is a generator of the multiplicative subgroup of the subfield $\operatorname{GF}(2^m)$

I don't understand why it holds that $\alpha^{2^{m} + 1}$ is a generator of the subfield?
Since you haven't heard of splitting fields, here is an argument based on more elementary group theory.

Let $G$ denote a cyclic group. Then, it is easy to show (or you might already know it) that if $\alpha \in G$ is an element of order $n$ (we write $\mathsf{ord}(\alpha)=n$ to denote this), then 
$$\mathsf{ord}(\alpha^k)= \frac{\mathsf{ord}(\alpha)}{\gcd(\mathsf{ord}(\alpha),k)} = \frac{n}{\gcd(n,k)}.$$
Choose $G = \text{GF}(2^{2m})^*$ whose generator $\alpha$ is of order $n = 2^{2m}-1$, observe that $2^{2m}-1 = (2^m-1)(2^m+1)$, and deduce that
$$\mathsf{ord}(\alpha^{2^{m}+1}) = \frac{2^{2m}-1}{\gcd(2^{2m}-1,2^{m}+1)} = \frac{(2^m-1)(2^m+1)}{2^{m}+1}
= 2^m-1.$$
Thus, $\alpha^{2^{m}+1}$ generates the unique cyclic subgroup (of order $2^m-1$) of $\text{GF}(2^{2m})^*$ which itself is of order $2^{2m}-1$. Now, $\text{GF}(2^{m})$ is a subfield of $\text{GF}(2^{2m})$ and so this subgroup must be $\text{GF}(2^{m})^*$.