This is Exercise 245 of the book "Fundamentals of Error-Correcting Codes" by W. C. Huffman and V. Pless, page 145.
Show that a monic irreducible reversible polynomial of degree greater than 1 cannot be a primitive polynomial except for the polynomial $1+x+x^2$ over $\mathbb{F}_2$.
Some background: The base field is assumed to be finite. An irreducible polynomial $f$ over a finite field is called primitive if its zeros are generators of the cyclic multiplicative group of the splitting field of $f$. The reciprocal of a polynomial $f$ is $x^{\operatorname{deg} f} f(1/x)$, which is the polynomial that arises from $f$ by reversing the sequence of coefficients. $f$ is called reversive if it agrees with its reciprocal.
The exercise is stated right after the definition of the notion "reciprocal" and "reversible", before stating elementary properties like that the zeros of the reciprocal are the inverses of the zeros of $f$. So I guess there should be an easy solution, but so far I fail to see it.
Apparently, writing helps for thinking, as I found this solution now:
Let $f\in\mathbb{F}_q[X]$ be reversible of degree $d$ and $\alpha$ a zero of $f$. The zeros of $f$ are then $\alpha, \alpha^q, \alpha^{q^2}, \ldots, \alpha^{q^{d-1}}$. The reciprocal of $f$ has the zero $\alpha^{-1}$. So if $f$ is reversible, $\alpha^{-1} = \alpha^{q^i}$ with $i\in\{0,\ldots,d-1\}$, or equivalenty $\alpha^{q^i + 1} = 1$. If $f$ is primitive, $\alpha$ has multiplicative order $q^d - 1$. This implies $q^i + 1 \geq q^d - 1$ or equivalently $q^i(q^{d-i} - 1) \leq 2$. With $d\geq 2$ and $i\leq d-1$, this forces $q=2$, $d=2$ and $i=1$. There is only one irreducible polynomial of degree $2$ over $\mathbb{F}_2$, namely $x^2 + x + 1$.
There is still the question if there is an easier solution without using the knwoledge that $\alpha^{-1}$ is a root of the reciprocal. Because in the book, that statement about the roots is found only after the excercise.