When is $3136 - 6272 n + 3136 n^2 + n^{14} =z^2$ for $z \in \mathbb N$?
Here was my approach: Consider $3136 - 6272 n + 3136 n^2 + n^{14} \pmod 5$. Since any integral square must be equivalent to 0, 1, or 4, modulo 5, we can rule out $n=5k+3$, since, making the substitution, we get
$$3136 - 6272 n + 3136 n^2 + n^{14} = 12544 + 62720 k + 78400 k^2 + (3 + 5 k)^{14}$$
which is always equivalent to 3 mod 5. Therefore $x \neq 5k+3$.
Continuing, we know a square must be equivalent to 0, 1, 3, or 4, modulo 6. However, the polynomial is equivalent to 2 and 5 modulo 6 whenever $n$ is equivalent to 2 or 5 modulo 6 respectively.
Hence, $n \neq 5k + 3, n \neq 6k+1, n \neq 6k+4, \cdots$
I suppose we can eliminate many arithmetic progressions from the possibilities of $n$, but that doesn't solve the case of whether the polynomial evaluated at some large prime for example might give output to a square integer. I'm not exactly sure how else to proceed here - any thoughts?
It may be checked that the expression
$$3136 - 6272n + 3136n^2 + n^{14} > (n^7+1)^2$$
reduces to
$2 \le n \le 3$. This can be seen by analyzing derivatives and critical points of both functions numerically. Then, since $3136 - 6272n + 3136n^2 > 0$ for $n>0$, we have that $n^{14} < 3136 - 6272n + 3136n^2 + n^{14}$.
It follows that for $n>3$,
$$u^2 < 3136 - 6272n + 3136n^2 + n^{14} < (u+1)^2$$
our expression lies strictly between two consecutive perfect squares, and is thus not a perfect square herself. Here $u=n^7$.