When $n=2$, we know 4 Fermat primes (among the 5 which are known) satisfying this condition:
$$ 1^2+2^2 =5$$ $$ 1^4+2^4 =17$$ $$ 1^8+2^8 = 257$$ $$ 1^{16}+2^{16} = 65537.$$
One may wonder for what other values of $(n,k)$, it holds that $\sum_{i=1}^n i^{2k}$ is a prime number.
Apparently, when $n=5$ and $2k= 1440$, \begin{align*} \sum_{i=1}^5 i^{1440}&=32870494973745590489672618520147226309471918340144695384282781773232645934767374502515598230671247791137527930976112524088008871699095171820372590939626862942685233205917502232958206871013324680536478277482222732171441694215728644181003593159822253865262588786239180084214018426223815623365507773026858373360823443708140883611306592330902932994635589466928339291445175747183614303780853567119243706761062506654051875550472877972022914140455264123266196392819668763698772234814545138902251784786312525104750087996929330185562414784278046396853711290578794116661635417485831969776614488207040722193777994839584756032261597855392326049625939136446894123910001325456787924317732389996265611844035119303216257160839794599676231991399494717047707966247305837634999211674025286390932779274786302608986436901299610556101262201055350271221229232970792516169435984584890424504786508517368097919942459145192988140070892281836785524154439529178872700448867894407573012070924416033111999815571623367458593539025033893379 \end{align*} is a $1007$ digits prime number.
Anyone knows another one? (with $n \gt 2$ and $k\gt0$)
I am also interested in any necessary conditions on $n$ and $k$ for $\sum_{i=1}^n i^{2k}$ to be prime. Clearly neither $n$ nor $n+1$ can be a multiple of $4$.
More generally, if $q$ is an odd prime number such that $q-1$ divides $2k$, $n$ cannot be $ 0\bmod {q^2} $ otherwise $$\sum_{i=1}^n i^{2k}\equiv \sum_{i\le n ,\mathrm {gcd}(i,q)=1} 1 = n-\lfloor {\frac{n}{q}}\rfloor \equiv 0 \pmod q $$ and cannot be a prime number. Then $n$ cannot be a multiple of $9$.
Also if $q$ is an odd prime number such that $q-1$ divides $2k$, $n+1$ cannot be $0 \bmod {q^2} $ either, otherwise $$\sum_{i=1}^n i^{2k}=\sum_{i=1}^{n+1} i^{2k}-(n+1)^{2k}\equiv \sum_{i\le n+1 ,\mathrm {gcd}(i,q)=1} 1 = n+1-\lfloor {\frac{n+1}{q}}\rfloor \equiv 0 \pmod q $$ and cannot be a prime number. Then $n+1$ cannot be a multiple of $9$ either.
And also if $q$ is an odd prime number such that $q-1$ divides $2k$, $2n+1$ cannot be $0 \bmod {q^2} $ either, otherwise
$$\sum_{i=1}^{n} i^{2k}=\sum_{i=1}^{2n+1} i^{2k}-\sum_{i=n+1}^{2n+1} i^{2k}=\sum_{i=1}^{2n+1} i^{2k}-\sum_{i=0}^{n} (2n+1-i)^{2k}$$
$$\sum_{i=1}^{n} i^{2k} \equiv \sum_{\underset{i\le 2n+1 }{\mathrm {gcd}(i,q)=1}} 1- \sum_{i=1}^{n} i^{2k}\equiv 2n+1-\lfloor {\frac{2n+1}{q}}\rfloor - n +\lfloor {\frac{n}{q}}\rfloor \equiv - n +\lfloor {\frac{n}{q}}\rfloor\pmod q.$$
But neither $n$ nor $n+1$ is multiple of $q$, otherwise both would be (since $q$ divides $2n+1$) which is impossible since $n$ and $n+1$ are coprime. Then $\lfloor {\frac{n}{q}}\rfloor=\lfloor {\frac{n+1}{q}}\rfloor$, then
$$\sum_{i=1}^{n} i^{2k} \equiv - n +\lfloor {\frac{n}{q}} \rfloor\equiv 2n+1-n+\lfloor {\frac{n+1}{q}} \rfloor\equiv n+1+\lfloor {\frac{n+1}{q}} \rfloor\pmod q,$$ then
$$2\sum_{i=1}^{n} i^{2k} \equiv 1 +\lfloor {\frac{n}{q}} \rfloor+\lfloor {\frac{n+1}{q}} \rfloor\pmod q.$$
But
$$\frac{2n+1}{q}-1=\lfloor {\frac{2n+1}{q}} \rfloor-1 \le \lfloor {\frac{n}{q}} \rfloor+\lfloor {\frac{n+1}{q}} \rfloor \le \lfloor {\frac{2n+1}{q}} \rfloor= \frac{2n+1}{q} $$
and $$\lfloor {\frac{n}{q}} \rfloor+\lfloor {\frac{n+1}{q}}\rfloor \neq \frac{2n+1}{q} $$ then
$$\frac{2n+1}{q}-1= \lfloor {\frac{n}{q}} \rfloor+\lfloor {\frac{n+1}{q}} \rfloor $$
$$2\sum_{i=1}^{n} i^{2k} \equiv \frac{2n+1}{q}\equiv 0 \pmod q$$ $$\sum_{i=1}^{n} i^{2k} \equiv 0 \pmod q$$ and cannot be a prime number.
Then $2n+1$ cannot be a multiple of $9$ either.
So we have shown that $n(n+1)(2n+1)$ cannot be divisible by $4$ nor by $9$ and I can show that a more general necessary condition is that $n(n+1)(2n+1)$ is squarefree. The proof is hereafter in the Appendix.
I also wonder whether another necessary condition is that the prime divisors of $2k$ are smaller than or equal to those of $n$. But we only have $1440=2^5\cdot3^2\cdot 5$ and the Fermat primes to support this.
Appendix
Let $\mathrm{rad}(n)$ be the product of the distinct prime factors of $n$. The purpose of this appendix is to show that when $k\ge1$, we have $$\sum_{i=1}^n j^{2k} \equiv 0 \bmod {\frac{n(n+1)(2n+1)}{\mathrm{rad} \big(n(n+1)(2n+1)\big)}}$$ whence if $n(n+1)(2n+1)$ is not squarefree, then $\sum_{i=1}^n j^{2k}$ is composite.
We first observe that it suffices to show that $S_{2k}(n):=\sum_{j=1}^nj^{2k}$ is divisible by $\frac{n}{\mathrm{rad}(n)}$. Indeed $n$ and $n+1$ are coprime and $S_{2k}(n)= S_{2k}(n+1)-(n+1)^{2k}$ is then also divisible by $\frac{n+1}{\mathrm{rad}(n+1)}$ , since $(n+1)^{2k}$ is divisible by $(n+1)^{2}$, hence by $\frac{n+1)}{\mathrm{rad}(n+1)}$ . Also $n$ and $2n+1$ are coprime and $S_{2k}(n)= S_{2k}(2n+1)-\sum_{i=n+1}^{2n+1}i^{2k}=S_{2k}(2n+1)-\sum_{i=0}^{n}(2n+1-i)^{2k}$, then $$S_{2k}(n)\equiv -\sum_{i=1}^{n}i^{2k}= -S_{2k}(n)\bmod {\frac{2n+1}{\mathrm{rad}(2n+1)}} $$ then $$2S_{2k}(n)\equiv 0\bmod {\frac{2n+1}{\mathrm{rad}(2n+1)}} $$ then $$S_{2k}(n)\equiv 0\bmod {\frac{2n+1}{\mathrm{rad}(2n+1)}} $$
Let $q$ be a prime divisor of $n$. By @reuns remark in the comment, $\sum_{j=1}^{n} j^{2k}\equiv \frac{n}{q}\sum_{j=1}^{q-1}j^{2k} \bmod q$ and this clearly shows that $\sum_{j=1}^{n} j^{2k}\equiv 0 \bmod q$ when $q^2$ divides $n$. $\square$
