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As the title says, I'm interested in integer solutions of the equation $(b^2+1)(c^2+1)=a^2+1$. Is it possible to parametrize the solutions as in the case of Pythagorean triples? If yes, then how would one proceed to find a parametrization in this case?

As suggested in the comments, the Pythagorean quadruples could be a good idea: we find that

$c = \frac{p^2-n^2-m^2}{2mp},\ b = \frac{p^2-n^2-m^2}{2np},\ a = \frac{(p^2+n^2+m^2)(p^2-n^2-m^2)}{4mnp^2}$ where $m,n,p$ are coprime and all the fraction are integers... I don't know if we can eliminate one of the $m,n,p$ using this...

Another take, which shows that there are infinitely many solutions: for fixed $b$ we can write $a^2-(b^2+1)c^2=b^2$ which is a Pell equation. This has a solution $a=b,c=0$ so it has infinitely many... Parametrizations, algorithms are available...

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    There is at least on non trivial solution (I mean except $(a,b,c)=(0,0,0)$) : take $(a,b,c)=(3,4,13)$. Also $(a,b,c)=(4,5,21)$ – 2017-02-15
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    I can generate some solutions using bruteforce and it seems there are many of them. – 2017-02-15
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    @Beni: Have you generated any that don't have $c=b+1$? – 2017-02-15
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    @JonasMeyer: For example $b=2,\ c = 14930352,\ a = 33385282$ – 2017-02-15
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    It appears a necessary condition is that $(a^2-b^2)(b^2+1)$ is also square. (with $a$ and $b$ as in the question, not Adren's comments) – 2017-02-15
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    Yes, that's $(a^2-b^2)(b^2+1)=(b^2+1)^2c^2$. – 2017-02-15
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    Possibly helpful: $(b,c,bc,a)$ form a "Pythagorean quadruple", and these are studied: http://ijpam.eu/contents/2007-35-3/10/10.pdf Although they don't have a full parametrization perhaps the extra condition (one number is a product of other two) combined with their results helps – 2017-02-15

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$(n^2+1)((n+1)^2 +1) = (n(n+1)+1)^2 + 1$. You can check this by simple expansion.

Hence, the cases such as $(3,4,13)$, $(4,5,21)$ etc. will come under this.

This is only an example. As I will get more I will keep updating this answer.

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Partial answer :

Let $n\in\mathbb{N}$ and :

$$(a,b,c)=(n,n+1,n^2+n+1)$$

We have :

$$(a^2+1)(b^2+1)=(n^2+1)(n^2+2n+2)=n^4+2n^3+3n^2+2n+2=(n^2+n+1)^2+1$$

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Show how the General formula allows us to solve this equation.

$$(a^2+1)(b^2+1)=c^2+1$$

We write this equation differently.

$$c^2-(a^2+1)b^2=a^2$$

Will do the replacement. $c=q+(k-1)b$

Then the equation takes the form.

$$q^2+2(k-1)qb+(k^2-2k-a^2)b^2=a^2$$

Now you can use the General formula. http://www.artofproblemsolving.com/community/c3046h1048219

Let the root equal to 1. This means that using the solutions of the equation Pell.

$$k^2-2a^2=1$$ Knowing the first solution. $$(k_0 ; a_0) - (3 ; 2)$$ You can find the following solution to the formula. $$k=3k_0+4a_0$$

$$a=2k_0+3a_0$$

Using the General formula and the solution of the equation Pell.

$$p^2-(a^2+1)s^2=\pm1$$

You can write the solution of the equation in this form.

$$c=kp^2\pm2(a^2+1)ps+k(a^2+1)s^2$$

$$b=p^2\pm2kps+(a^2+1)s^2$$