$$ n^4-4n^3+22n^2-36n+18 = (n-1)^4 +16(n-1)^2+1 = [(n-1)^2+8]^2 -63 $$Let this be equal to $k^2$: $$ [(n-1)^2+8]^2 - k^2 = 63 \Rightarrow [(n-1)^2+8+k][(n-1)^2+8-k] = 63 = (63,1), (21,3), (9,7) $$Solving, $n = 1 \pm 2\sqrt{6}, -1, 3, 1$. The only positive integers are $\boxed{n = 1,3}$