Convert boolean indices to integer with nonzero

[ricardoV94]

Description

It seems to be faster, both in the C and Numba backends, and regardless of whether the idx is constant or symbolic:

import pytensor
import pytensor.tensor as pt

x = pt.vector("x", shape=(10_000,))
idx = np.random.default_rng(1).binomial(n=1, p=0.5, size=x.type.shape).astype(bool)
fn1 = pytensor.function([x], x[idx], trust_input=True)
fn1.dprint()

fn2 = pytensor.function([x], x[idx.nonzero()], trust_input=True)
fn2.dprint()

x_test = np.arange(x.type.shape[0]).astype(x.dtype)
%timeit fn1(x_test)
%timeit fn2(x_test)

# AdvancedSubtensor [id A] 0
#  ├─ x [id B]
#  └─ [ True  Tr ... lse False] [id C]
# AdvancedSubtensor1 [id A] 0
#  ├─ x [id B]
#  └─ [   0    1 ... 9994 9996] [id C]
# 52.5 μs ± 1.09 μs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
# 17.1 μs ± 494 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

Difference is also large in the numba backend. This would allow us to simplify the codebase quite a lot by getting rid of boolean indices in our graph representation. There's only one case where boolean indices are not equivalent to .nonzero(), which is when the boolean variable is scalar, but we don't support that explicitly anyway.