Case 1: $p=2,3$ $p=3\implies (p-3)^p+p^2=9$ which is perfect square. $p=2\implies (p-3)^p+p^2=5$ which is not perfect square. Case 2: $p\geq 5$ +) $3\nmid (p-3)^p+p^2\implies(p-3)^p+p^2\equiv 1\pmod{3}$ since it is perfect square. $\implies (p-3)^p\equiv 1-p^2\equiv 0\pmod{3}\implies 3\mid (p-3)^p\implies 3\mid p$ which is absurd. +) $3\mid (p-3)^p+p^2\implies 3\mid p^p+p^2\implies 3\mid p^p+1\implies p\equiv 2\pmod{3}$ Since $p\neq 2$, $p=6k+5\implies (p-3)^p+p^2=(6k+2)^{6k+5}+(36k^2+60k+25)$. By Euler, $(p-3)^p+p^2\equiv (6k+2)^5+6k-2\stackrel{\text{binomial}}{\equiv} (480k+32)+6k-2\equiv 3\pmod{9}$. But note that $(p-3)^p+p^2$ can't be perfect square since $v_3((p-3)^p+p^2)=1$. Thus, $p=3$.