I checked using MATLAB that $x$ is a primitive element of $\text{GF}(2^m)$ for $m\le 16$. Is the statement true for $m > 16$?
EDIT: In this question, we represent an element of $\text{GF}(2^m)$ as a polynomial of $x$ with degree at most $m-1$ with coefficients drawn from $\text{GF}(2)$. I used the default primitive polynomials (see https://www.mathworks.com/help/comm/ref/gf.html ).
Short answer: if the irreducible polynomial used to generate the binary extension field $\mathrm{GF}(2^m)$ is primitive, then yes $x$ is always a primitive element. Although, the irreducible polynomial need not be primitive. Matlab always uses a primitive polynomial as its default irreducible polynomial.
A Galois extension field $\mathrm{GF}(p^m)$ is created using an irreducible polynomial over $\mathrm{GF}(p)$ of degree $m$. If the irreducible polynomial is primitive, then the field element $x$ is a primitive element $\alpha$ of the field, meaning the field elements can be represented by its powers $\mathrm{GF}(p^m) = \{0, 1, \alpha, \alpha^2, \dots, \alpha^{p^m - 2}\}$. When the irreducible polynomial is primitive, the polynomial representation of field elements use $\alpha$ as the polynomial indeterminate (since $x = \alpha$), $c_{m-1}\alpha^{m-1} + \dots + c_1\alpha + c_0 \in \mathrm{GF}(p^m)$ for $c_i \in \mathrm{GF}(p)$.
Definition: A polynomial $f(x) \in \mathrm{GF}(p)[x]$ is reducible over $\mathrm{GF}(p)$ if it can be represented as $f(x) = g(x)h(x)$ for some non-constant $g(x), h(x) \in \mathrm{GF}(p)[x]$ of strictly lower degree. If $f(x)$ is not reducible, it is said to be irreducible. Since Galois fields are not algebraically closed, such irreducible polynomials exist.
Definition: A degree-$m$ polynomial $f(x)$ over $\mathrm{GF}(p)$ is primitive if it is irreducible and $f(x)\ |\ (x^k - 1)$ for $k = p^m - 1$ and no $k$ less than $p^m - 1$.
I created a python package galois that extends numpy arrays for Galois fields. It mimics a lot of functionality of Matlab's finite field functions. Here are some examples using the library.
In $\mathrm{GF}(2^4)$, which I'm using because it's a small field, $x^4 + x + 1$ and $x^4 + x^3 + 1$ are primitive polynomials of degree 4. Matlab and my library's default irreducible polynomial is $x^4 + x + 1$ (which is primitive). My library contains a
repr_table()method to display all field element representations as powers of a primitive element and as polynomials over $\mathrm{GF}(p)$. Since the irreducible polynomial is primitive, notice the polynomial $x$ = $\alpha$ and all the powers of $\alpha$ uniquely generate the field.However, there are irreducible polynomials of degree 4 over $\mathrm{GF}(2)$ that are not primitive, namely $x^4 + x^3 + x^2 + x + 1$. Constructing $\mathrm{GF}(2^4)$ using a non-primitive polynomial does not result in $x$ being a primitive element. In this case, $\alpha = x + 1$ is the minimal primitive element of $\mathrm{GF}(2^4)$, although there are others. Notice that the polynomial $x = \alpha^{12}$.
We can also display all the powers of the element $\beta = x$ and see that $\beta$ does not generate the field. Specifically, $\textrm{ord}(\beta) = 5 \ne 2^4 - 1$.
The most common example of a Galois field that does not use a primitive polynomial is the Galois field used in AES. AES uses $\mathrm{GF}(2^8)$ with irreducible polynomial $x^8 + x^4 + x^3 + x + 1$, which is not primitive.
Hope this helps!