Consider first the case $x,y<0$. Subtract $1980z$ from both sides of the equation. Since
$1975^{4^{30}}+2010-1980z$
is an integer, $19^x+5^y$ has to be as well. But for $x$ and $y$ negative, these are fractions, each of them smaller than $1/2$ since $19>2$ and $5>2$:
$0<19^x + 5^y = \frac{1}{19^{-x}}+\frac{1}{5^{-y}}\leq \frac{1}{19}+\frac{1}{5} < 1$
So it is impossible that this is an integer.
Now consider $x,y\geq 0$. Since $19$ is odd, $19^x$ is odd as well. The same holds for $5^y$, while $1980z$ is even for all $z$.
So the left hand side of the equation is odd + odd + even, which is even.
Can you do the same for the right hand side? You will get a contradiction.