Recall that factoring is the process of taking a sum/difference (things added/subtracted)
and renaming it as a product (things multiplied).
An expression of the form
$\,A^2  B^2\,$
is called a difference of squares.
It's a difference, because the last operation being performed is subtraction.
It's a difference of squares, because both $\,A^2\,$ and $\,B^2\,$ are squares.
Using FOIL:
$\,\cssId{s11}{(A + B)(A  B)}
\cssId{s12}{= A^2  AB + AB  B^2}
\cssId{s13}{= A^2  B^2}\,$
Thus, we have the following result:
FACTORING A DIFFERENCE OF SQUARES
For all real numbers
$\,A\,$ and $\,B\,$:
$$
\cssId{s17}{A^2  B^2 = (A + B)(A  B)}
$$
EXAMPLES:
Factor:
$x^2  4$
Solution:
$\cssId{s22}{x^2  4}
\cssId{s23}{= x^2  2^2}
\cssId{s24}{= (x + 2)(x  2)}
$
Factor:
$9  x^2$
Solution:
$\cssId{s28}{9  x^2}
\cssId{s29}{= 3^2  x^2}
\cssId{s30}{= (3 + x)(3  x)}
$
Factor:
$4x^2  9$
Solution:
$\cssId{s34}{4x^2  9}
\cssId{s35}{= (2x)^2  3^2}
\cssId{s36}{= (2x + 3)(2x  3)}
$
Factor:
$49x^2  64y^2$
Solution:
$\cssId{s40}{49x^2  64y^2}
\cssId{s41}{= (7x)^2  (8y)^2}
\cssId{s42}{= (7x + 8y)(7x  8y)}
$
Factor:
$x^6  25$
Solution:
$\cssId{s46}{x^6  25}
\cssId{s47}{= (x^3)^2  5^2}
\cssId{s48}{= (x^3 + 5)(x^3  5)}
$
Factor:
$x^2  5$
Solution:
Since $\,5\,$ is not a perfect square, this cannot be factored using integers.
Note that it can be factored if we're allowed to use nonintegers:
$\cssId{s54}{x^2  5}
\cssId{s55}{= x^2  (\sqrt{5})^2}
\cssId{s56}{= (x + \sqrt{5})(x  \sqrt{5})}
$
In this exercise, you are factoring over the integers.
That is, you are to use only the integers for your factoring.
Recall that the integers are:
$\,\{\ldots,3,2,1,0,1,2,3,\ldots\}$
Factor:
$x^2 + 4$
Solution:
This can't be factored using integers.
Usually, a sum of squares can't be factored.
(☆ The remaining discussion is beyond the scope of Algebra I;
it is included for the benefit of more advanced readers.)
The expression $\,x^2 + 4\,$ can't even be factored using real numbers.
It can be factored if we're allowed to use numbers that aren't real:
$\cssId{s69}{x^2 + 4}
\cssId{s70}{= x^2  (2i)^2}
\cssId{s71}{= (x + 2i)(x  2i)}\,$,
where $i^2 = 1$
In general, a sum of squares can't be factored.
However, a sum of squares might also be a sum of cubes, which is factorable, like this:
$x^6 + 64$


$\ \ \cssId{s76}{= (x^3)^2 + 8^2}$

(so, it's a sum of squares) 
$\ \ \cssId{s78}{= (x^2)^3 + 4^3}$

(it's also a sum of cubes, which can be factored) 
$\ \ \cssId{s80}{= (x^2+4)(x^4  4x^2 + 16)}$

(use this: $\,A^3 + B^3 = (A + B)(A^2  AB + B^2)\ \ $) 
So, you can't just make a blanket statement that sums of squares aren't factorable.