Find all primes $p$ such that $ p^3-4p+9 $ is a perfect square.

If $p^3-4p+9=n^2$ then $n^2\equiv 9 \pmod{p}$ and $\pm n \equiv 3 \pmod{p}$ since $n^2-9\equiv 0$ can only have two roots modulo a prime.

By choosing the appropriate sign we can write $$ \begin{align} n^2=(-n)^2 &= (kp+3)^2 \\ p^3-4p+9 &= k^2p^2+6kp+9 \\ 0 &= p^2-k^2p-(6k+4) \\ p & = \frac{k^2\pm \sqrt{k^4+24k+16}}{2} \end{align} $$ where we used $p\ne 0$ since it's a prime. This can only have an integer solution for $p$ when $k^4+24k+16$ is a square (as @tkr suggested in the comments). But $$ \begin{array}{ll} (k^2-1)^2 = k^4-2k^2+1 < k^4+24k+16<(k^2)^2 & \text{if }k<-11 \\ (k^2)^2 < k^4+24k+16< k^4+2k^2+1 =(k^2+1)^2 & \text{if }k>12 \end{array} $$ so if $k<-11$ or $k>12$ then $k^4+24k+16$ lies between two squares and hence is not a square. This leaves 24 values to check, which you could do by hand. It leads to solutions when $k=-3,0,3$ which give integral values for $p$ of $-2,2,7,11$, of which only $2,7,11$ would normally be considered prime.