We are given: $$\gcd(a,b)^2+\mathrm{lcm}(a,b)^2+a^2b^2=2020^k$$Since it is well-known that $\gcd(a,b)\mathrm{lcm}(a,b)=ab$, this is equivalent to $$(\gcd(a,b)^2+1)(\mathrm{lcm}(a,b)^2+1)=2020^k+1$$Suppose $k$ is odd, then taking modulo $43$, we have $(\gcd(a,b)^2+1)(\mathrm{lcm}(a,b)^2+1)\equiv 0\pmod{43}$, thus either $\gcd(a,b)^2\equiv -1\pmod{43}$ or $\mathrm{lcm}(a,b)^2\equiv -1\pmod{43}$, but since $43\equiv 3 \pmod 4$, then $-1$ is not a quadratic residue modulo $43$, therefore our assumption was wrong, hence $k$ must be an even number.

Let $x=\gcd(a,b)$ and $y=\text{lcm}(a,b)$. We know that $xy=ab$, so we can get \[x^2+y^2+(xy)^2=2020^k \implies (x^2+1)(y^2+1)=2020^{k}+1.\]Now suppose $k$ is odd. If so, then we have \[2021|2020^k+1=(x^2+1)(y^2+1) \implies 43|(x^2+1)(y^2+1).\]This implies that $x^2\equiv -1\pmod{43}$ or $y^2\equiv -1\pmod{43}$, but since $43\equiv 3\pmod 4$, $\left(\frac{-1}{43}\right)=-1$, so we have a contradiction. Thus $k$ must be even. sad sniped