Find a prime that divides $14^7+14^2+1$ As the title says we seek to

Find a prime that divides $14^7+14^2+1$

There is a caveat though. This was part of a contest for high-school students so undergraduate Number Theory tools such as modular arithmetic should be avoided if we wish to remain true to the spirit of the competition.
That said, I did verify-employing exactly such tools-that if $p$ is the said prime then $p\neq2,3,5,7$.
Probably some form of simplifying the expression is required to solve it but it eludes me.
EDIT
The initial formulation of this question was "find the smallest such prime". This is highly unlikely to be achieved by pen and paper on the timeframe of a contest. For more details on why one can look at @lulu's answer below. 
As noted on the comments by @JyrkiLahtonen, a similar question was posted before. 
See here as well for answers. 
 A: The reason we can get one factor immediately is that $7 \equiv 1 \pmod 3.$ As a result, both nontrivial roots of unity are roots of $x^7 + x^2 + 1,$ meaning that $(x - \omega)(x- \omega^2) = x^2 + x + 1$ must divide the polynomial $x^7 + x^2 + 1.$ Here $\omega^3 = 1$ but $\omega \neq 1$ 
If this seems uncomfortable, just consider that $x^2 + x + 1$ is the minimal polynomial for $\omega$ over $\mathbb Q,$ and must divide any polynomial for which $\omega$ is a root. Furthermore, the Gauss theorem on content tells us that the quotient polynomial has integer coefficients, not just rational.
Similar: the polynomial $x^{141} + x^{93} + x^{82} + x^{44} + 1$ is divisible by $x^4 + x^3 + x^2 + x + 1,$ consider a fifth root of unity.
A: We can factor the polynomial $$x^7+x^2+1=(x^2+x+1)\times (x^5-x^4+x^2-x+1)$$
Letting $x=14$ shows that $211$ is a prime factor.  You still have to prove it is the least prime factor, but at least you can work with a smaller number.
A: $$x^7+x^2+1=x^7-x+x^2+x+1=x(x^3-1)(x^3+1)+x^2+x+1=$$
$$=(x^2-x)(x^2+x+1)(x^3+1)+x^2+x+1=$$
$$=(x^2+x+1)(x^5-x^4+x^2-x+1).$$
