BarisKoyuncu wrote:

WLOG $a-b=p$ where $p$ is a prime number.
i) $p|b$
Let $b=pd$ . Then, $(a,b)=(pd+p,pd)=p(d+1,d)=p$ and $[a,b]=[pd+p,pd]=p[d+1,d]=pd(d+1)$ . Hence, $p+pd(d+1)=2021^c\Rightarrow p|2021^c\Rightarrow p|2021\Rightarrow p=43,47$ .
i.a) $p=43$
$2021^c=p(d^2+d+1)=43(d^2+d+1)\Rightarrow d^2+d+(1-43^{c-1}\cdot 47^c)=0$ . Hence, the number $\triangle_d 1-4(1-43^{c-1}\cdot 47^c)=4\cdot 43^{c-1}\cdot 47^c-3$ must a perfect square. But, $4\cdot 43^{c-1}\cdot 47^c-3\equiv -3\pmod{47}$ and $47\equiv 2\pmod{3}$ . So, this number cannot be a perfect square. Contradiction.
i.b) $p=47$
$2021^c=p(d^2+d+1)=47(d^2+d+1)\Rightarrow d^2+d+(1-43^c\cdot 47^{c-1})=0$ . Hence, the number $\triangle_d 1-4(1-43^c\cdot 47{c-1}c)=4\cdot 43^c\cdot 47^{c-1}-3$ must a perfect square. If $c\ge 2$ , again $4\cdot 43^c\cdot 47^{c-1}-3\equiv -3\pmod{47}$ . Contradiction. Thus, $c=1$ . Then, $\triangle_d=4\cdot 43^c\cdot 47^{c-1}-3=4\cdot 43-3=169=13^2$ . Hence, $d=\dfrac{-1\pm\sqrt{\triangle_d}}{2}=\dfrac{-1\pm 13}{2}=\{-7,6\}$ . Since $d>0$ , we find that $d=6$ . Then, $b=pd=47\cdot 6=282$ and $a=b+p=282+47=329\Rightarrow (a+b)^2+4=611^2+4\equiv 0\pmod{5}$ . Clearly, $611^2+4>5$ , so it is composite.
ii) $p\not |b$
Then, $(a,b)=(b+p,b)=(b,p)=1$ and $[a,b]=[b+p,b]=(b+p)b$ . Hence, $1+(b+p)b=2021^c\Rightarrow b^2+pb+(1-2021^c)=0$ . Hence, the number $\triangle_d=p^2-4(1-2021^c)=p^2+4\cdot 2021^c-4$ must be a perfect square. Let $p^2+4\cdot 2021^c-4=t^2$ where $t\in \mathbb{Z^+}$ . Then, $b=\dfrac{-p\pm \sqrt{t^2}}{2}$ and since $b>0$ , we find that $b=\dfrac{t-p}{2}$ . Then, $(a+b)^2+4=\left(\dfrac{t-p}{2}+\dfrac{t+p}{2}\right)^2+4=t^2+4$ . Suppose that $t^2+4=q$ where $q$ is a prime number. Hence, $p^2+4\cdot 2021^c=t^2+4=q$ .
ii.a) $c$ is even.
Let $c=2c_1$ . Then, $p^2+(2\cdot 2021^{c_1})^2=q=t^2+2^2$ . But, each prime number can be written in $1$ or $0$ different way as the sum of $2$ perfect squares. Thus, $\{p,2\cdot 2021^{c_1}\}=\{t,2\}$ . Clearly, $2\cdot 2021^{c_1}>2$ so $p=2$ . Then $q=p^2+4\cdot 2021^c\equiv 0\mod{2}\Rightarrow q=2$ but $p^2+4\cdot 2021^c>$ . Contradiction.
ii.b) $c$ is odd.
If $p\neq 3$ , then $t^2+4=p^2+4\cdot 2021^c\equiv (-1)^c\equiv -1\pmod{3}\Rightarrow t^2\equiv 2\pmod{3}$ . Contradiction. So $p=3$ . Then, $t^2+4=9+4\cdot 2021^c\equiv 9\pmod{47}\Rightarrow t^2\equiv 5\pmod{47}$ . Contradiction.

You can excise a fair amount of work here. Let $d={\rm gcd}(a,b)$ with $a=da_1$ and $b=db_1$ , $(a_1,b_1)=1$ . Assume w.l.o.g. $a_1>b_1$ (clearly $a\ne b$ ). We then have $d(a_1-b_1)=p$ for a prime $p$ , thus $d\in\{1,p\}$ . Now, if $d=p$ then we obtain that
$d\left(b_1^2+b_1+1\right)=43^c\cdot 47^c.$ If $47\mid b_1^2+b_1+1$ , then $47\mid (2b_1+1)^2+3$ , but $(-3/47)=-1$ as $47\equiv -1\pmod{6}$ . Hence, in this case, $47^c\mid d = p$ , thus $c=1$ and $d=47$ is the only possibility. With this we find $b_1^2+b_1+1=43$ , for which $(a_1,b_1)=(7,6)$ is obtained; and for this solution, $(a+b)^2+4>5$ is divisible by $5$ , hence is the conclusion.

This brings us to the case $(a,b)=1$ , $a-b=p$ , which is handled exactly as demonstrated by Baris. (Let me also add that one way to prove the also contradiction, $t^2\equiv 5\pmod{47}$ , in the very last step is to use the quadratic reciprocity: $(5/47)(47/5)=1$ whereas $(57/5)=(2/5)=-1$ .)