jmc

We could subtract $292$ to the other side and try to solve the sixth-degree equation, but that would be ugly and we have no guarantee that it would even work. We notice that we can add a $x^4-x^4+x^2-x^2$ to the polynomial without changing its value: $$x^6-2x^5+(x^4-x^4)+2x^3+(x^2-x^2)-2x+1=292.$$ We regroup the terms and factor the left side, leaving the $292$ alone. $$\begin{array}{rl}(x^6-2x^5+x^4)+(-x^4+2x^3+-x^2)+(x^2-2x+1)& =292\\ x^4(x^2-2x+1)-x^2(x^2-2x+1)+1(x^2-2x+1)& =292\\ (x^2-2x+1)(x^4-x^2+1)& =292\\ (x-1{)}^2(x^4-x^2+1)& =292.\end{array}$$ To see a different way of obtaining this factorization, we could also group the $x^6$ and $1$ terms together and factor, giving: $$\begin{array}{rl}(x^6+1)+(-2x^5+2x^3-2x)& =292\\ (x^2+1)(x^4-x^2+1)-2x(x^4-x^2+1)& =292\\ (x^4-x^2+1)(x^2+1-2x)& =292\\ (x^4-x^2+1)(x-1{)}^2& =292.\end{array}$$ Since $x$ is an integer, then $x^4-x^2+1$ and $x-1$ are integers, so they must be factors of $292$. The prime factorization of $292$ is $2^2\cdot 73$. The value $(x-1{)}^2$ must be a square which divides $292$, and we can see that the only squares which divide $292$ are $1$ and $4$.

If $(x-1{)}^2=1$, then $x-1=\pm 1$, and $x=2$ or $x=0$. If $x=0$, it's easy to see from the original equation that it won't work, since the left side of the original equation would be $1$ while the right side would be $292$. If $x=2$, then the left side would equal $(2^4-2^2+1)(2-1{)}^2=(16-4+1)(1{)}^2=13\ne 292$. So both values of $x$ are impossible.

Thus $(x-1{)}^2=4$, so $x-1=\pm 2$ and $x=3$ or $x=-1$. If $x=-1$, then we have the left side as $((-1{)}^4-(-1{)}^2+1)((-1)-1{)}^2=(1-1+1)(-2{)}^2=1(4)=4\ne 292$. If $x=3$, then we have $(3^4-3^2+1)(3-1{)}^2=(81-9+1)(2^2)=73\cdot 2^2=292$ as desired. Thus the only possibility for $x$ is $\boxed{3}$.