If $n$ is odd, the expression is $3\pmod 4$, so $n$ is even. Let $n = 2k$. Then $9^k + 4k^2 + 2019$ is a perfect square. Therefore, $9^k + 4k^2 + 2019\geq (3^k + 1)^2 = 9^k + 2\cdot 3^k + 1$. So $4k^2 + 2018\geq 2\cdot 3^k\rightarrow k\leq 6$. Computation gives that $4$ is the only solution. In order to reduce computation, note that $k\neq 5$ since the expression is then $3\pmod 5$, and $k\neq 6$, since then the expression divides $3$ but not $9$.

CWMI 2019 P1 wrote: Determine all the possible positive integer $n,$ such that $3^n+n^2+2019$ is a perfect square. Solution: Firstly, $n=1,2$ give no solutions. $$3^n+n^2+2019=k^2$$Notice: $k^2 \equiv n^2 \equiv 0,1 \pmod{3}$. If $0 \pmod{3}$, then $2 \le v_3(n^2), v_3 (k^2)$, but $v_3(2019)=1$ $\implies$ contradiction! Hence, $k^2$ $\equiv$ $n^2$ $\equiv$ $1$ $\pmod{3}$. Next, taking $\pmod{4}$ $\implies$ $(-1)^n+n^2+3$ $\equiv$ $k^2 \pmod{4}$. If $n$ is odd, gives contradiction! hence, $n,k$ are even. $$3^n=(3^{n/2})^2<3^n+n^2+2019<3^n+2 \times 3^{n/2} < (3^{n/2}+1)^2$$Which is true for $n \geq 14$. Checking, for $3 \le n \le 13$ $\implies$ $\boxed{n=4} \qquad \blacksquare$

If $n$ is odd, the expression is $3\pmod 4$, so $n$ is even. Let $n = 2k$. Then $9^k + 4k^2 + 2019$ is a perfect square. Therefore, $9^k + 4k^2 + 2019\geq (3^k + 1)^2 = 9^k + 2\cdot 3^k + 1$. So $4k^2 + 2018\geq 2\cdot 3^k\rightarrow k\leq 6$. Computation gives that $4$ is the only solution. In order to reduce computation, note that $k\neq 5$ since the expression is then $3\pmod 5$, and $k\neq 6$, since then the expression divides $3$ but not $9$.

Take mod $4$ gives you $n$ even, so let $n=2k$. Now the expression becomes $9^k+4k^2+2019=x^2$ As we have the expression being larger than $9^k$, it can’t be square if it’s smaller than $(3^k+1)^2$. So we have $2\cdot 3^k+1\leq 4k^2+2019$, so $k<7$. $k=1,4$ fails by checking. $k=3,6$ failed by computing $v_3$, $k=5$ fails by mod $5$, and $k=2$ works.

CWMI 2019 P1 wrote: Determine all the possible positive integer $n,$ such that $3^n+n^2+2019$ is a perfect square. Solution: Firstly, $n=1,2$ give no solutions. $$3^n+n^2+2019=k^2$$Notice: $k^2 \equiv n^2 \equiv 0,1 \pmod{3}$. If $0 \pmod{3}$, then $2 \le v_3(n^2), v_3 (k^2)$, but $v_3(2019)=1$ $\implies$ contradiction! Hence, $k^2$ $\equiv$ $n^2$ $\equiv$ $1$ $\pmod{3}$. Next, taking $\pmod{4}$ $\implies$ $(-1)^n+n^2+3$ $\equiv$ $k^2 \pmod{4}$. If $n$ is odd, gives contradiction! hence, $n,k$ are even. $$3^n=(3^{n/2})^2<3^n+n^2+2019<3^n+2 \times 3^{n/2} < (3^{n/2}+1)^2$$Which is true for $n \geq 14$. Checking, for $3 \le n \le 13$ $\implies$ $\boxed{n=4} \qquad \blacksquare$