Show that there are infinitely many natural numbers such that $a^2+b^2=c^2+3 .$

There are infinitely many even numbers. Let us assume that $b$ is even, as doing so does not limit the number of possible solutions. We can rearrange the original equation to $$c^2-a^2=b^2-3$$ where $b^2-3$ must be odd.

It is well know that the sum of the first $n$ odd numbers equals $n^2$. Consequently, every odd number is the difference of two consecutive squares. It may also be the difference of other pairs of squares, but it is at least the difference of two consecutive squares.

Therefore, every odd number $b^2-3$ can be represented in at least one way as the difference of two squares.

For example, choose $b=6$ (an even number), so $b^2-3=33$.

$33$ is the $17$th odd number, and the sum of the first $17$ odd numbers is $17^2=289$. The $17$th odd number succeeds the $16$th odd number, and the sum of the first $16$ odd numbers is $16^2=256$. Plainly, $289-256=33=b^2-3$. It is also the case that $49-16=33$ (which is one of OP's original instances), but that is not necessary to establish that the number of solutions is infinite.

Added by edit: In my answer as originally posted, I addressed only the question as to whether there are infinite solutions. That garnered a query by OP as to whether all solutions can be found. The answer to that is yes, but the formulation is somewhat involved.

Based on the fact that $\sum_{i=1}^k(2i-1)=k^2$ we can show that any odd number is the difference of two squares: $n=r^2-s^2$

The formula $(\sum_{i=1}^{\frac{n+1}{2}}(2i-1))-(\sum_{i=1}^{\frac{n-1}{2}}(2i-1))$ represents the difference of two consecutive squares. Each sum contains the same terms, which by subtraction cancel, except the first sum contains an extra term corresponding to $2(\frac{n+1}{2})-1$ which simply equals $n$. In order that the indices on the sums be integers, it is required that $n$ be odd, but beyond that, any odd number affords a solution. This more fully explains the conclusion presented in my first answer.

If $n$ has integer factors, a larger set of solutions is available. Let $t$ be a factor of $n$. Then $$n=\Biggl(\sum_{i=1}^{\frac{\frac{n}{t}+t}{2}}(2i-1)\Biggr)-\Biggl(\sum_{i=1}^{\frac{\frac{n}{t}-t}{2}}(2i-1)\Biggr)$$ with the following constraints: $\frac{n}{t}>t$ so that the indices are greater than $0$ and $\frac{n}{t}\equiv t \bmod 2$ so that the indices are integers. The first sum is the square of the superior index $$\Bigl(\frac{\frac{n}{t}+t}{2}\Bigr)^2=\frac{(\frac{n}{t})^2+2n+t^2}{4}$$ The second sum is the square of the superior index $$\Bigl(\frac{\frac{n}{t}-t}{2}\Bigr)^2=\frac{(\frac{n}{t})^2-2n+t^2}{4}$$ Subtracting the second from the first yields $\frac{4n}{4}=n$

So when $n$ has appropriate factors, it can be represented in additional ways as the difference of two squares.

In the context of the question, we can set $n=b^2-3$. It is not necessary to write out all of the terms and indices explicitly here; the principles have been demonstrated.

In the example I used, $b=6$, the resulting $b^2-3=33$ is divisible by $3$. this affords indices $\frac{\frac{33}{3}\pm 3}{2}=7,4$ yielding $7^2-4^2=33$. Since the other factor of $33$ is $11$ and $\frac{33}{11}\not > 11$, this factor does not satisfy the constraints, and we have found all solutions corresponding to $b=6$.

Brainstorming from doing something similar with pythagorean triples.

If $b^2 = 2a + 1 + 3=2a+4$ then $a^2 + b^2 = a^2 + 2a + 1 + 3 = (a+1)^2 + 3$.

If $b$ is any even number, $2k; k > 1$ then if $a = \frac {(2k)^2-4}2= 2k^2-2$

And if $c = 2k^2 -1$ then

$a^2 + b^2 = $

$(2k^2 -2)^2 + 4k^2 =$

$4k^4 - 8k^2 + 4 + 4k^2 =$

$4k^4 - 4k^2 + 1 + 3 =$

$(2k^2 - 1)^2 + 3=$

$c^2 + 3$.

Yep.... that seems to do it.

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Perhaps more formally I should have done

$a^2 + b^2 = c^2 + 3 \implies$

$c^2 - a^2 = b^2 -3$

$(c-a)(c+a) = b^2 -3$

If I let $d = c-a$ be an odd number and $e = d+2a = c+a$ be a larger odd number then so long as we have

$b^2 -3 = d*e$ is an odd number then we have a solutions.

So if $b$ is any even number $> 2$ then $b^2 -3$ is odd number. If we let $b^2 -3 = d*e; d< e$ be any factors of $b^2-3$ (if $b^2 -3$ is prime or a prime square we can let $d = 1$ and $e=b^2 -3$). Then we let $a = \frac {e-d}2$ and $c = \frac {e+d}2$ we then have

$b^2 -3 = d*e = (c-a)(c+a) = c^2 - a^2$ and

$a^2 +b^2 = c^2 + 3$.

...

If $b$ is odd and $b^2 -3$ is even then $b= 2k+1$ and $b^2 -3 = 4k^2 + 4k -2$ and is divisible by $2$ but not $4$. $c-a$ and $c+a$ must both be the same parity so this will not be possible.