Does this polynomial evaluate to prime number whenever $x$ is a natural number? I am trying to prove or disprove following statment:

$x^2-31x+257$ evaluates to a prime number whenever $x$ is a natural number.

First of all, I realized that  we can't factorize this  polynomial  using its square root like
 $$ax^2+bx+c=a(x-x_1)(x-x_2)$$
because discriminant is negative, also I used excel for numbers from 1 to 1530 (just for checking), and yes it gives me prime numbers, unfortunately I dont know  what the formula is, which evaluates  every natural number for given input, maybe we may write
$x_{k+1}=k+1$     for $k=0,\ldots\infty$, but how can I use  this  recursive statment?
I have tried  instead of  $x$, put $k+1$ in this polynomial, and so I got
$$(k+1)^2-31k+257=k^2+2k+1-31k+257=k^2-29k+258$$
but now this last  polynomial for $k=1$ evaluates to $259-29=230$, which is not prime, but are my steps correct? Please help me.

If $x$ is divisible by $257$ then so is $x^2 - 31 x + 257$. 
More generally, if $f(x)$ is any polynomial  $f(x)$ is divisible by $f(y)$ whenever $x-y$ is divisible by $f(y)$.  So there are no non-constant polynomials that produce primes for all positive integers.  
