Output a primitive element for each field size

Mathematica, 127 bytes

Do[For[i=2*2^n,PolynomialMod[x^Divisors[2^n-1]+1,i~IntegerDigits~2~FromDigits~x,Modulus->2]~Count~0!=1,i--];Print@{2,i},{n,32}]

Explanation:

If we choose a primitive polynomial as the irreducible polynomial, then \$x\$ is a primitive element. A irreducible polynomial of degree \$n\$ is primitive if and only if it has order \$2^n - 1\$, i.e., it is divisible by \$x^{2^n - 1} - 1\$, but not divisible by any \$x^i - 1\$, where \$i\$ runs through all the proper divisors of \$2^n - 1\$.

Output:

Outputs integers whose bits are the coefficients of the polynomials. For example, \$8589934581\$ is \$111111111111111111111111111110101\$ when written in binary, so it represents the polynomial $$x^{32}+x^{31}+x^{30}+x^{29}+x^{28}+x^{27}+x^{26}+x^{25}+\\x^{24}+x^{23}+x^{22}+x^{21}+x^{20}+x^{19}+x^{18}+x^{17}+\\x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^9+\\x^8+x^7+x^6+x^5+x^4+x^2+1$$

{2,3}

{2,7}

{2,13}

{2,25}

{2,61}

{2,115}

{2,253}

{2,501}

{2,1019}

{2,2041}

{2,4073}

{2,8137}

{2,16381}

{2,32743}

{2,65533}

{2,131053}

{2,262127}

{2,524263}

{2,1048531}

{2,2097145}

{2,4194227}

{2,8388589}

{2,16777213}

{2,33554351}

{2,67108849}

{2,134217697}

{2,268435427}

{2,536870805}

{2,1073741801}

{2,2147483533}

{2,4294967287}

{2,8589934581}

Pari/GP, 114 bytes

Inspired by isaacg's answer in another question.

for(n=1,32,for(i=1,2^n,if(sumdiv(2^n-1,d,Mod(x,f=Mod(Pol(binary(2*2^n-i)),2))^d==1)==1,print([x,lift(f)]);break)))

Try it online!

---

If built-ins are allowed:

Pari/GP, 61 bytes (non-competing)

for(i=1,32,print([ffprimroot(ffgen(f=ffinit(2,i))),lift(f)]))

Try it online!