Why is $a^2+b^2$ more likely to be prime than $a^2+b$?

Taking a random $a$ and $b$, the probability that $a^2+b^2$ is a multiple of $3$ is about $1/9$, since this happens only if $a$ and $b$ are both multiples of $3$. But the probability of $a^2+b$ being divisible by $3$ is about $1/3$, since no matter what $a$ is there is a roughly $1/3$ chance that $b\equiv -a^2\pmod 3$.

You will get a similar effect for any prime $p\equiv 3\pmod 4$, since for these primes $-1$ is not a quadratic residue, which implies there is no solution to $a^2+b^2\equiv 0\pmod p$ except $a\equiv 0,b\equiv 0$.

Say $1\leq a, b\leq N$, then $$ a^2+b\leq N^2+N\qquad\text{and}\qquad a^2+b^2\leq2N^2. $$ But while every prime $p$ is of the form $a^2+b$ (just take $a=1$, $b=p-1$) only those congruent to $1$ modulo $p$ are of the form $a^2+b^2$.

Thus by the prime number theorem (which says that there are about $x/\log x$ primes $\leq x$) the probability to hit a prime for random values of $a^2+b$ is $$ p_1=\frac{\log(N^2+N)}{N^2+N}=\frac{\log(N)+\log(N+1)}{N(N+1)} $$ while the probability of hitting a prime of the form $a^2+b^2$ is $$ p_2=\frac12\frac{\log(2N^2)}{2N^2}=\frac12\frac{\log(2)+2\log(N)}{2N^2} $$ where the factor $\frac12$ is there because the primes $\equiv1\bmod4$ have density $\frac12$.

Note that $p_1>p_2$.