jmc

We need some basic facts from number theory: $a^0=1$ for any $a,$ $1^b=1$ for any $b,$ and $(-1{)}^c=1$ if $c$ is an even integer. Unless the base is a complex number (which is excluded since we are looking for integer solutions), there are no other ways to get an RHS of $1.$ Thus, either the exponent is zero $($giving the equation $25-x^2=0),$ the base is $1$ $($giving $x-2=1),$ or the base is $-1$ and the exponent is even $($giving the simultaneous equations $x-2=-1$ and $25-x^2=2n$ for some integer $n).$ Solving the first equation gives $x=\pm 5,$ and solving the second gives $x=3.$ The third equation implies that $x=1,$ in which case $25-x^2=24$ is indeed even, so $x=1$ is a valid solution. In all, there are $\boxed{4}$ integer solutions.