Let's find those integer values of $x$ for which substituting into the expression $x^{2}+28 x+889$ yields the square of an integer.
Let $p$ be an integer such that

$$
x^{2}+28 x+889=(x+14)^{2}+693=p^{2}
$$

that is,

$$
p^{2}-(x+14)^{2}=(p-x-14)(p+x+14)=693
$$

Since $p-x$ and $p+x$ are integers, both factors must be integers, so we can only get solutions from the two-factor decompositions of 693. If $693=a \cdot b$, then by equating the factors in pairs,

$$
p-x-14=a, \quad p+x+14=b
$$

from which

$$
p=\frac{a+b}{2} \quad \text { and } \quad x=\frac{b-a}{2}-14
$$

Since $a$ and $b$ can only be odd, $p$ and $x$ will certainly be integers. We will certainly not get a prime for $p$ if $a$ and $b$ have a common divisor, because if they do, it will also be odd, and will remain after division by 2. The relatively prime pairs of $a, b$ that we need to check, to see if their sum is prime, and calculate $x$ only in the affirmative cases, are as follows:

Since $693=3^{2} \cdot 7 \cdot 11$ as a product of powers of different prime numbers, the two 3 factors must remain together (either in $a$ or in $b$), so based on the form $693=9 \cdot 7 \cdot 11$, the $a, b$ divisor pairs to be examined are:

$$
\begin{aligned}
& a, b=1,693, \text { half of their sum } 347, \text { prime, } \\
& a, b=9, \quad 77, \text { half of their sum } 43, \text { prime, } \\
& a, b=7, \quad 99, \text { half of their sum } 53, \text { prime, } \\
& a, b=11,63, \text { half of their sum } 37, \text { prime, }
\end{aligned}
$$

thus we get solutions from all four decompositions, and two from each. From the first $a, b$ pair,

$$
x=\frac{693-1}{2}-14=332 \quad \text { and } \quad x=\frac{1-693}{2}-14=-360
$$

and similarly from the others,

$$
x=20 \text { and }-48, \quad 32 \text { and }-60, \quad 8 \text { and }-40
$$

finally, listing them in ascending order:

$$
x=-360,-60,-48,-40,8,20,32,332
$$

are the sought values.